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Geometry: Prove that two angles are not equal
Clash Royale CLAN TAG#URR8PPP
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(This is just a question for fun. I saw a commercial logo today and I was inspired.
I have posted answers for this question and you may post alternative answers!)
Question
In the figure, $triangle ABC$ is half of a square and $M$ is the midpoint of $BC$. Prove that $alphaneqbeta$.
Solution
$triangle ABM$ and $triangle AMC$ have the same area. They have a
common side $AM$.
Note that the area of either triangle is given by
$S=frac12(AB)(AM)sinalpha=frac12(AC)(AM)sinbeta$.
But $ABneq AC$. So the equality holds only if $alphaneqbeta$.
geometry alternative-proof
edited Jul 21 at 18:10
Serge Seredenko
1621315
asked Jul 21 at 16:43
Mythomorphic
5,0961732
add a comment |Â
up vote
15
down vote
favorite
(This is just a question for fun. I saw a commercial logo today and I was inspired.
I have posted answers for this question and you may post alternative answers!)
Question
In the figure, $triangle ABC$ is half of a square and $M$ is the midpoint of $BC$. Prove that $alphaneqbeta$.
Solution
$triangle ABM$ and $triangle AMC$ have the same area. They have a
common side $AM$.
Note that the area of either triangle is given by
$S=frac12(AB)(AM)sinalpha=frac12(AC)(AM)sinbeta$.
But $ABneq AC$. So the equality holds only if $alphaneqbeta$.
geometry alternative-proof
edited Jul 21 at 18:10
Serge Seredenko
1621315
asked Jul 21 at 16:43
Mythomorphic
5,0961732
8
This is an illustration that the tangent function is nonlinear.
– Yves Daoust
Jul 21 at 17:03
1
Fun question! Thanks for sharing :)
– Sambo
Jul 21 at 17:09
add a comment |Â
up vote
15
down vote
favorite
up vote
15
down vote
favorite
(This is just a question for fun. I saw a commercial logo today and I was inspired.
I have posted answers for this question and you may post alternative answers!)
Question
In the figure, $triangle ABC$ is half of a square and $M$ is the midpoint of $BC$. Prove that $alphaneqbeta$.
Solution
$triangle ABM$ and $triangle AMC$ have the same area. They have a
common side $AM$.
Note that the area of either triangle is given by
$S=frac12(AB)(AM)sinalpha=frac12(AC)(AM)sinbeta$.
But $ABneq AC$. So the equality holds only if $alphaneqbeta$.
geometry alternative-proof
edited Jul 21 at 18:10
Serge Seredenko
1621315
asked Jul 21 at 16:43
Mythomorphic
5,0961732
(This is just a question for fun. I saw a commercial logo today and I was inspired.
I have posted answers for this question and you may post alternative answers!)
Question
In the figure, $triangle ABC$ is half of a square and $M$ is the midpoint of $BC$. Prove that $alphaneqbeta$.
Solution
$triangle ABM$ and $triangle AMC$ have the same area. They have a
common side $AM$.
Note that the area of either triangle is given by
$S=frac12(AB)(AM)sinalpha=frac12(AC)(AM)sinbeta$.
But $ABneq AC$. So the equality holds only if $alphaneqbeta$.
geometry alternative-proof
edited Jul 21 at 18:10
Serge Seredenko
1621315
asked Jul 21 at 16:43
Mythomorphic
5,0961732
edited Jul 21 at 18:10
Serge Seredenko
1621315
edited Jul 21 at 18:10
Serge Seredenko
1621315
edited Jul 21 at 18:10
Serge Seredenko
1621315
1621315
asked Jul 21 at 16:43
Mythomorphic
5,0961732
asked Jul 21 at 16:43
Mythomorphic
5,0961732
asked Jul 21 at 16:43
Mythomorphic
5,0961732
5,0961732
8
This is an illustration that the tangent function is nonlinear.
– Yves Daoust
Jul 21 at 17:03
1
Fun question! Thanks for sharing :)
– Sambo
Jul 21 at 17:09
add a comment |Â
8
This is an illustration that the tangent function is nonlinear.
– Yves Daoust
Jul 21 at 17:03
1
Fun question! Thanks for sharing :)
– Sambo
Jul 21 at 17:09
8
8
This is an illustration that the tangent function is nonlinear.
– Yves Daoust
Jul 21 at 17:03
This is an illustration that the tangent function is nonlinear.
– Yves Daoust
Jul 21 at 17:03
1
1
Fun question! Thanks for sharing :)
– Sambo
Jul 21 at 17:09
Fun question! Thanks for sharing :)
– Sambo
Jul 21 at 17:09
add a comment |Â
13 Answers
13
active
oldest
votes
up vote
20
down vote
accepted
By contradiction: if the angles were equal, then $BM:MC=AB:AC$ (angle bisector theorem) and as a consequence $BM>MC$, which is false.
answered Jul 21 at 16:59
Aretino
21.7k21342
add a comment |Â
up vote
11
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A mostly visual proof:
Construct a second square with edge $BC$ as shown. Since $AC=BD$, $CM=BM$, and $angle ACM$ and $angle DBM$ are right angles, $triangle ACM cong triangle DBM$.
Therefore $angle CMA cong angle BMD$, so $A$, $M$, and $D$ are collinear. And $mangle BDM = mangle CAM = beta$.
Now if $alpha = beta$, then $triangle ABD$ is isosceles, with $AB = BD$. But this is plainly false since $AB = BD sqrt2$. So $alpha neq beta$.
answered Jul 22 at 2:44
aschepler
1,50379
add a comment |Â
up vote
9
down vote
$hspace5cm$
If $alpha=beta$, then:
$$1=tan 45^circ=tan(alpha+beta)=tan(2beta)=frac2tanbeta1-tan^2beta=frac2cdot frac121-left(frac12right)^2=frac43,$$
hence a contradiction. So, $alphane beta$.
answered Jul 22 at 5:31
farruhota
13.7k2632
Nice! Seems that with your method we can generalize the question when $BC$ is divided into $n$ equal segments.
– Mythomorphic
Jul 22 at 7:07
Nice observations both on commercial logo and on generalization!
– farruhota
Jul 22 at 12:10
Haha, thank you! Not an advertisement, but the name of the company is Convoy. Take a look of it lol
– Mythomorphic
Jul 22 at 12:15
I believe you should clarify that you're calculating tan β by dividing the edge sizes. Because for a moment it seemed that you replaced tan(22.5°) with 1/2 which is false :)
– Pedro A
Jul 22 at 21:20
add a comment |Â
up vote
6
down vote
Construct the circumcircle of $triangle AMC$ and translate $triangle BMA$ such that $BM$ coincides with $MC$. If the two angles were equal, $A'$ (the third vertex of the translated triangle) would lie on the circle, but it does not. Thus the two angles are not equal.
edited Jul 21 at 17:45
mvw
30.4k22250
answered Jul 21 at 16:56
Parcly Taxel
33.6k136588
Parcly.Very nice.
– Peter Szilas
Jul 21 at 19:12
add a comment |Â
up vote
4
down vote
Let $C'$ be the image of the orthogonal projection of $M$ onto $AB$. Suppose contrary that $alpha=beta$. Then, $MCA$ and $MC'A$ are congruent triangles. Thus, $MB=MC'$ and there is a contradiction here, which I will leave it as a mystery. In fact, you can use a similar argument to show that $alpha<beta$.
answered Jul 21 at 16:51
Batominovski
23.3k22777
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3
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Let line MD be constructed parallel to AC, with D on AB, so angle DMA = MAC. Now DM < DA, because DM < BD (Pythagorean Th.). But in any triangle, a greater side lies opposite a greater angle. Therefore angle DMA (that is, MAC) > BAM.
edited Jul 22 at 20:09
mvw
30.4k22250
answered Jul 21 at 22:50
Timothy Smith
1313
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2
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From the diagram we have
$$
alpha + beta = pi/4
$$
If $alpha = beta$ this would give $alpha = beta = pi / 8$.
Again from the diagram: Let the side of the square be $x$, then
$$
tan beta = (x/2) / x = 1/2
$$
However
$$
tan(pi/8) = sqrt2-1 < 1/2
$$
So $beta ne pi/8$ and $a ne beta$.
Appendix:
The addition theorem for the tangent yields
$$
tan(2x) = frac2tan(x)1-tan^2(x)
$$
So
$$
underbracetan(2cdot pi/8)_1 = frac2tan(pi/8)1-tan^2(pi/8)
$$
and we have the quadratic equation in $u = tan(pi/8)$:
$$
beginalign
1-u^2 &= 2 u iff \
1 &= u^2 + 2u iff \
2 &= u^2 + 2u + 1 = (u+1)^2 iff \
u &= pmsqrt2-1
endalign
$$
where we need the positive solution.
answered Jul 21 at 16:51
mvw
30.4k22250
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up vote
2
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Drop a perpendicular from $M$ to the line $AB,$ meeting $AB$ at $P.$
That is, $P$ is the point on line $AB$ closest to $M.$
Since $angle ABM$ is not a right angle, $P$ is not $B,$
and therefore $MP < BM.$ Since $BM = CM,$ then $MP < CM.$
If $alpha = beta$ then the two right triangles
$triangle AMC$ and $triangle AMP$ would be congruent,
since all three angles and one side ($AM$) would be congruent.
But since $MP < CM,$ the triangles are not congruent.
Hence $alpha neq beta.$
answered Jul 21 at 17:38
David K
48.2k340107
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up vote
1
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By the law of sines,
$$sqrt2fracsin(alpha)MB=frac1MA=fracsin(beta)MC.$$ Since $MC=MB$, we conclude that $sqrt2sin(alpha)=sin(beta)$. Clearly neither $alpha$ nor $beta$ are integer multiples of $pi$, so $alphaneqbeta$.
answered Jul 21 at 20:47
Alex S
17.6k11959
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up vote
1
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First assume that $alpha$ = $beta$ and that both would be equal to $fracpi8^circ$ (since $alpha = beta$ and $alpha + beta = fracpi4 to 2beta = fracpi4 to beta = fracpi8$).
If we call side $overlineAC$, n, then side $overlineMC$ is $fracn2$. Next, we solve for $overlineAM$ in terms of n to show that $alpha = beta$ for all values n.
$overlineAM^2 = n^2 + (fracn2)^2 = n^2 + fracn^24 = frac4n^2 + n^24 = frac5n^24$
$overlineAM = sqrtfrac5n^24 = fracsqrt5n^22 = fracnsqrt52$ by the pythagorean theorem
Now that we have $overlineAM$, we can find $cosbeta$.
$cosbeta=fracnoverlineAM = fracnfracnsqrt52 = frac2nnsqrt5 = frac2sqrt5 = frac2sqrt55$
$cosbeta = frac2sqrt55$
We have that $cosbeta = frac2sqrt55$ and we already established that $beta = fracpi8$.
This would mean that $cosfracpi8 = frac2sqrt55$. We can find $cosfracpi8 = fracsqrt2 + sqrt22$, which implies that $fracsqrt2 + sqrt22 = frac2sqrt55$.
If we square both sides, we get that $frac2+sqrt24 = frac2025 = frac45$. Then we can split the fraction, simplify, then subtract and simplify.
$frac24 + fracsqrt24 = frac45$
$frac12 + fracsqrt24 = frac45$
$fracsqrt24 = frac45 - frac12 $
$fracsqrt24 = frac310$
Next we can multiply both sides by 4 and conclude that...
$sqrt2 = frac1210 = frac65$
This is impossible since it has been proven that $sqrt2$ is irrational and can't be written as a ratio of 2 integers. This would mean that $sqrt2 neq frac65$ or $fracsqrt2 + sqrt22 neq frac2sqrt55$ or $cosfracpi8neq frac2sqrt55$
This means that $beta neq fracpi8$. Since $alpha + beta = fracpi4$, $beta$ would have to be equal to $fracpi8$, but we proved that it cant be. This means that $alpha neq beta$ since $beta$ can no longer be multiplied by 2 to be $fracpi4$
This method works with tangent and sine as well (I believe it's a lot less work if I had used tangent instead of cosine, but this was how I first solved it)
I'm a high-school student and this is my first response, feedback is always appreciated!
answered Jul 22 at 15:18
Jamar Sullivan
112
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1
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Here's one intuitive way to analyze the problem:
If you're standing on point $A$ and looking at a car driving at constant speed on a straight line $BC$, you'll have to turn your head very fast when the car is closest to you (on point $C$) and slow the head rotation as the car drives away.
answered Jul 22 at 20:41
Eric Duminil
1,6961516
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0
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The figure $triangle ABC$ is the half of a square therefore the sides $overlineAB$ and $overlineBC$ are of equal length. Now consider the triangle $triangle ABM$. It is a rectangular triangle such as the original one.
Therefore the tangens of the angle $alpha$ is given by
$$tan(alpha)~=~fracoverlineBMoverlineAB$$
hence $M$ is the middle of the side $overlineBC$ this simplifies to
$$tan(alpha)~=~fracoverlineBMoverlineAB~=~fracfracoverlineBC2overlineAB~=~fracfracoverlineBC2overlineBC~=~frac12\
alpha~=~arctanleft(frac12right)$$
Now go back to the triangle $triangle ABC$. The tangens of the angle $alpha + beta$ is given by
$$tan(alpha + beta)~=~fracoverlineBCoverlineAB$$
which equals $1$. The tangens of the sum of two different angles is given by
$$tan(x+y)~=~fractan(x)+tan(y)1-tan(x)tan(y)$$
with $x=alpha$,$y=beta$ and $tan(alpha)=frac12$ you get
$$beginalign
tan(alpha+beta)~=~1~=~fracfrac12+tan(beta)1-frac12tan(beta)~iff~1-frac12tan(beta)~&=~frac12+tan(beta)\
tan(beta)&=frac13\
beta&=arctanleft(frac13right)
endalign$$
Hence $arctanleft(frac12right)neqarctanleft(frac13right)$ and so $alphaneqbeta$.
answered Jul 21 at 17:13
mrtaurho
700219
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0
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Assume $alpha$ and $beta$ are both equal to $22.5^circ$ (sum to forty-five).
The angle $angle ABM$ must equal $45^circ$ because it bisects the square.
The angle $angle ACM$ occupies a square corner, so it must equal $90^circ$.
$alpha + angle BMA$ must not sum to more than $135^circ$ ($180^circ$ $-$ $45^circ$ from $angle ABM$).
If alpha is equal to beta and both equal $22.5^circ$, then BMA
must equal $112.5$ degrees ($135^circ - 22.5^circ$).
BMA + AMC must equal less than $180$ degrees (one hemicircle); if $angle BMA = 112.5^circ$, then $angle AMC$ must be $67.5^circ$.
Given $beta = alpha = 22.5^circ$, $angle AMC + angle ACM$ must equal $157.5^circ$ ($180$ minus $beta$); given the same assumption, $angle AMC$ is equal to $67.5^circ$ and sums with $angle ACM$ to $147.5^circ$.
Alpha must not be equal to beta.
answered Jul 23 at 3:16
Paul Ferris
33
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13 Answers
13
active
oldest
votes
13 Answers
13
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
20
down vote
accepted
By contradiction: if the angles were equal, then $BM:MC=AB:AC$ (angle bisector theorem) and as a consequence $BM>MC$, which is false.
answered Jul 21 at 16:59
Aretino
21.7k21342
add a comment |Â
up vote
20
down vote
accepted
By contradiction: if the angles were equal, then $BM:MC=AB:AC$ (angle bisector theorem) and as a consequence $BM>MC$, which is false.
answered Jul 21 at 16:59
Aretino
21.7k21342
add a comment |Â
up vote
20
down vote
accepted
up vote
20
down vote
accepted
By contradiction: if the angles were equal, then $BM:MC=AB:AC$ (angle bisector theorem) and as a consequence $BM>MC$, which is false.
answered Jul 21 at 16:59
Aretino
21.7k21342
By contradiction: if the angles were equal, then $BM:MC=AB:AC$ (angle bisector theorem) and as a consequence $BM>MC$, which is false.
answered Jul 21 at 16:59
Aretino
21.7k21342
answered Jul 21 at 16:59
Aretino
21.7k21342
answered Jul 21 at 16:59
Aretino
21.7k21342
answered Jul 21 at 16:59
Aretino
21.7k21342
21.7k21342
add a comment |Â
add a comment |Â
up vote
11
down vote
A mostly visual proof:
Construct a second square with edge $BC$ as shown. Since $AC=BD$, $CM=BM$, and $angle ACM$ and $angle DBM$ are right angles, $triangle ACM cong triangle DBM$.
Therefore $angle CMA cong angle BMD$, so $A$, $M$, and $D$ are collinear. And $mangle BDM = mangle CAM = beta$.
Now if $alpha = beta$, then $triangle ABD$ is isosceles, with $AB = BD$. But this is plainly false since $AB = BD sqrt2$. So $alpha neq beta$.
answered Jul 22 at 2:44
aschepler
1,50379
add a comment |Â
up vote
11
down vote
A mostly visual proof:
Construct a second square with edge $BC$ as shown. Since $AC=BD$, $CM=BM$, and $angle ACM$ and $angle DBM$ are right angles, $triangle ACM cong triangle DBM$.
Therefore $angle CMA cong angle BMD$, so $A$, $M$, and $D$ are collinear. And $mangle BDM = mangle CAM = beta$.
Now if $alpha = beta$, then $triangle ABD$ is isosceles, with $AB = BD$. But this is plainly false since $AB = BD sqrt2$. So $alpha neq beta$.
answered Jul 22 at 2:44
aschepler
1,50379
add a comment |Â
up vote
11
down vote
up vote
11
down vote
A mostly visual proof:
Construct a second square with edge $BC$ as shown. Since $AC=BD$, $CM=BM$, and $angle ACM$ and $angle DBM$ are right angles, $triangle ACM cong triangle DBM$.
Therefore $angle CMA cong angle BMD$, so $A$, $M$, and $D$ are collinear. And $mangle BDM = mangle CAM = beta$.
Now if $alpha = beta$, then $triangle ABD$ is isosceles, with $AB = BD$. But this is plainly false since $AB = BD sqrt2$. So $alpha neq beta$.
answered Jul 22 at 2:44
aschepler
1,50379
A mostly visual proof:
Construct a second square with edge $BC$ as shown. Since $AC=BD$, $CM=BM$, and $angle ACM$ and $angle DBM$ are right angles, $triangle ACM cong triangle DBM$.
Therefore $angle CMA cong angle BMD$, so $A$, $M$, and $D$ are collinear. And $mangle BDM = mangle CAM = beta$.
Now if $alpha = beta$, then $triangle ABD$ is isosceles, with $AB = BD$. But this is plainly false since $AB = BD sqrt2$. So $alpha neq beta$.
answered Jul 22 at 2:44
aschepler
1,50379
answered Jul 22 at 2:44
aschepler
1,50379
answered Jul 22 at 2:44
aschepler
1,50379
answered Jul 22 at 2:44
aschepler
1,50379
1,50379
add a comment |Â
add a comment |Â
up vote
9
down vote
$hspace5cm$
If $alpha=beta$, then:
$$1=tan 45^circ=tan(alpha+beta)=tan(2beta)=frac2tanbeta1-tan^2beta=frac2cdot frac121-left(frac12right)^2=frac43,$$
hence a contradiction. So, $alphane beta$.
answered Jul 22 at 5:31
farruhota
13.7k2632
Nice! Seems that with your method we can generalize the question when $BC$ is divided into $n$ equal segments.
– Mythomorphic
Jul 22 at 7:07
Nice observations both on commercial logo and on generalization!
– farruhota
Jul 22 at 12:10
Haha, thank you! Not an advertisement, but the name of the company is Convoy. Take a look of it lol
– Mythomorphic
Jul 22 at 12:15
I believe you should clarify that you're calculating tan β by dividing the edge sizes. Because for a moment it seemed that you replaced tan(22.5°) with 1/2 which is false :)
– Pedro A
Jul 22 at 21:20
add a comment |Â
up vote
9
down vote
$hspace5cm$
If $alpha=beta$, then:
$$1=tan 45^circ=tan(alpha+beta)=tan(2beta)=frac2tanbeta1-tan^2beta=frac2cdot frac121-left(frac12right)^2=frac43,$$
hence a contradiction. So, $alphane beta$.
answered Jul 22 at 5:31
farruhota
13.7k2632
Nice! Seems that with your method we can generalize the question when $BC$ is divided into $n$ equal segments.
– Mythomorphic
Jul 22 at 7:07
Nice observations both on commercial logo and on generalization!
– farruhota
Jul 22 at 12:10
Haha, thank you! Not an advertisement, but the name of the company is Convoy. Take a look of it lol
– Mythomorphic
Jul 22 at 12:15
I believe you should clarify that you're calculating tan β by dividing the edge sizes. Because for a moment it seemed that you replaced tan(22.5°) with 1/2 which is false :)
– Pedro A
Jul 22 at 21:20
add a comment |Â
up vote
9
down vote
up vote
9
down vote
$hspace5cm$
If $alpha=beta$, then:
$$1=tan 45^circ=tan(alpha+beta)=tan(2beta)=frac2tanbeta1-tan^2beta=frac2cdot frac121-left(frac12right)^2=frac43,$$
hence a contradiction. So, $alphane beta$.
answered Jul 22 at 5:31
farruhota
13.7k2632
$hspace5cm$
If $alpha=beta$, then:
$$1=tan 45^circ=tan(alpha+beta)=tan(2beta)=frac2tanbeta1-tan^2beta=frac2cdot frac121-left(frac12right)^2=frac43,$$
hence a contradiction. So, $alphane beta$.
answered Jul 22 at 5:31
farruhota
13.7k2632
edited Jul 22 at 6:15
answered Jul 22 at 5:31
farruhota
13.7k2632
answered Jul 22 at 5:31
farruhota
13.7k2632
answered Jul 22 at 5:31
farruhota
13.7k2632
13.7k2632
Nice! Seems that with your method we can generalize the question when $BC$ is divided into $n$ equal segments.
– Mythomorphic
Jul 22 at 7:07
Nice observations both on commercial logo and on generalization!
– farruhota
Jul 22 at 12:10
Haha, thank you! Not an advertisement, but the name of the company is Convoy. Take a look of it lol
– Mythomorphic
Jul 22 at 12:15
I believe you should clarify that you're calculating tan β by dividing the edge sizes. Because for a moment it seemed that you replaced tan(22.5°) with 1/2 which is false :)
– Pedro A
Jul 22 at 21:20
add a comment |Â
Nice! Seems that with your method we can generalize the question when $BC$ is divided into $n$ equal segments.
– Mythomorphic
Jul 22 at 7:07
Nice observations both on commercial logo and on generalization!
– farruhota
Jul 22 at 12:10
Haha, thank you! Not an advertisement, but the name of the company is Convoy. Take a look of it lol
– Mythomorphic
Jul 22 at 12:15
I believe you should clarify that you're calculating tan β by dividing the edge sizes. Because for a moment it seemed that you replaced tan(22.5°) with 1/2 which is false :)
– Pedro A
Jul 22 at 21:20
Nice! Seems that with your method we can generalize the question when $BC$ is divided into $n$ equal segments.
– Mythomorphic
Jul 22 at 7:07
Nice! Seems that with your method we can generalize the question when $BC$ is divided into $n$ equal segments.
– Mythomorphic
Jul 22 at 7:07
Nice observations both on commercial logo and on generalization!
– farruhota
Jul 22 at 12:10
Nice observations both on commercial logo and on generalization!
– farruhota
Jul 22 at 12:10
Haha, thank you! Not an advertisement, but the name of the company is Convoy. Take a look of it lol
– Mythomorphic
Jul 22 at 12:15
Haha, thank you! Not an advertisement, but the name of the company is Convoy. Take a look of it lol
– Mythomorphic
Jul 22 at 12:15
I believe you should clarify that you're calculating tan β by dividing the edge sizes. Because for a moment it seemed that you replaced tan(22.5°) with 1/2 which is false :)
– Pedro A
Jul 22 at 21:20
I believe you should clarify that you're calculating tan β by dividing the edge sizes. Because for a moment it seemed that you replaced tan(22.5°) with 1/2 which is false :)
– Pedro A
Jul 22 at 21:20
add a comment |Â
up vote
6
down vote
Construct the circumcircle of $triangle AMC$ and translate $triangle BMA$ such that $BM$ coincides with $MC$. If the two angles were equal, $A'$ (the third vertex of the translated triangle) would lie on the circle, but it does not. Thus the two angles are not equal.
edited Jul 21 at 17:45
mvw
30.4k22250
answered Jul 21 at 16:56
Parcly Taxel
33.6k136588
Parcly.Very nice.
– Peter Szilas
Jul 21 at 19:12
add a comment |Â
up vote
6
down vote
Construct the circumcircle of $triangle AMC$ and translate $triangle BMA$ such that $BM$ coincides with $MC$. If the two angles were equal, $A'$ (the third vertex of the translated triangle) would lie on the circle, but it does not. Thus the two angles are not equal.
edited Jul 21 at 17:45
mvw
30.4k22250
answered Jul 21 at 16:56
Parcly Taxel
33.6k136588
Parcly.Very nice.
– Peter Szilas
Jul 21 at 19:12
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Construct the circumcircle of $triangle AMC$ and translate $triangle BMA$ such that $BM$ coincides with $MC$. If the two angles were equal, $A'$ (the third vertex of the translated triangle) would lie on the circle, but it does not. Thus the two angles are not equal.
edited Jul 21 at 17:45
mvw
30.4k22250
answered Jul 21 at 16:56
Parcly Taxel
33.6k136588
Construct the circumcircle of $triangle AMC$ and translate $triangle BMA$ such that $BM$ coincides with $MC$. If the two angles were equal, $A'$ (the third vertex of the translated triangle) would lie on the circle, but it does not. Thus the two angles are not equal.
edited Jul 21 at 17:45
mvw
30.4k22250
answered Jul 21 at 16:56
Parcly Taxel
33.6k136588
edited Jul 21 at 17:45
mvw
30.4k22250
edited Jul 21 at 17:45
mvw
30.4k22250
edited Jul 21 at 17:45
mvw
30.4k22250
30.4k22250
answered Jul 21 at 16:56
Parcly Taxel
33.6k136588
answered Jul 21 at 16:56
Parcly Taxel
33.6k136588
answered Jul 21 at 16:56
Parcly Taxel
33.6k136588
33.6k136588
Parcly.Very nice.
– Peter Szilas
Jul 21 at 19:12
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Parcly.Very nice.
– Peter Szilas
Jul 21 at 19:12
Parcly.Very nice.
– Peter Szilas
Jul 21 at 19:12
Parcly.Very nice.
– Peter Szilas
Jul 21 at 19:12
add a comment |Â
up vote
4
down vote
Let $C'$ be the image of the orthogonal projection of $M$ onto $AB$. Suppose contrary that $alpha=beta$. Then, $MCA$ and $MC'A$ are congruent triangles. Thus, $MB=MC'$ and there is a contradiction here, which I will leave it as a mystery. In fact, you can use a similar argument to show that $alpha<beta$.
answered Jul 21 at 16:51
Batominovski
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up vote
4
down vote
Let $C'$ be the image of the orthogonal projection of $M$ onto $AB$. Suppose contrary that $alpha=beta$. Then, $MCA$ and $MC'A$ are congruent triangles. Thus, $MB=MC'$ and there is a contradiction here, which I will leave it as a mystery. In fact, you can use a similar argument to show that $alpha<beta$.
answered Jul 21 at 16:51
Batominovski
23.3k22777
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up vote
4
down vote
up vote
4
down vote
Let $C'$ be the image of the orthogonal projection of $M$ onto $AB$. Suppose contrary that $alpha=beta$. Then, $MCA$ and $MC'A$ are congruent triangles. Thus, $MB=MC'$ and there is a contradiction here, which I will leave it as a mystery. In fact, you can use a similar argument to show that $alpha<beta$.
answered Jul 21 at 16:51
Batominovski
23.3k22777
Let $C'$ be the image of the orthogonal projection of $M$ onto $AB$. Suppose contrary that $alpha=beta$. Then, $MCA$ and $MC'A$ are congruent triangles. Thus, $MB=MC'$ and there is a contradiction here, which I will leave it as a mystery. In fact, you can use a similar argument to show that $alpha<beta$.
answered Jul 21 at 16:51
Batominovski
23.3k22777
edited Jul 21 at 17:14
answered Jul 21 at 16:51
Batominovski
23.3k22777
answered Jul 21 at 16:51
Batominovski
23.3k22777
answered Jul 21 at 16:51
Batominovski
23.3k22777
23.3k22777
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up vote
3
down vote
Let line MD be constructed parallel to AC, with D on AB, so angle DMA = MAC. Now DM < DA, because DM < BD (Pythagorean Th.). But in any triangle, a greater side lies opposite a greater angle. Therefore angle DMA (that is, MAC) > BAM.
edited Jul 22 at 20:09
mvw
30.4k22250
answered Jul 21 at 22:50
Timothy Smith
1313
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up vote
3
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Let line MD be constructed parallel to AC, with D on AB, so angle DMA = MAC. Now DM < DA, because DM < BD (Pythagorean Th.). But in any triangle, a greater side lies opposite a greater angle. Therefore angle DMA (that is, MAC) > BAM.
edited Jul 22 at 20:09
mvw
30.4k22250
answered Jul 21 at 22:50
Timothy Smith
1313
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let line MD be constructed parallel to AC, with D on AB, so angle DMA = MAC. Now DM < DA, because DM < BD (Pythagorean Th.). But in any triangle, a greater side lies opposite a greater angle. Therefore angle DMA (that is, MAC) > BAM.
edited Jul 22 at 20:09
mvw
30.4k22250
answered Jul 21 at 22:50
Timothy Smith
1313
Let line MD be constructed parallel to AC, with D on AB, so angle DMA = MAC. Now DM < DA, because DM < BD (Pythagorean Th.). But in any triangle, a greater side lies opposite a greater angle. Therefore angle DMA (that is, MAC) > BAM.
edited Jul 22 at 20:09
mvw
30.4k22250
answered Jul 21 at 22:50
Timothy Smith
1313
edited Jul 22 at 20:09
mvw
30.4k22250
edited Jul 22 at 20:09
mvw
30.4k22250
edited Jul 22 at 20:09
mvw
30.4k22250
30.4k22250
answered Jul 21 at 22:50
Timothy Smith
1313
answered Jul 21 at 22:50
Timothy Smith
1313
answered Jul 21 at 22:50
Timothy Smith
1313
1313
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up vote
2
down vote
From the diagram we have
$$
alpha + beta = pi/4
$$
If $alpha = beta$ this would give $alpha = beta = pi / 8$.
Again from the diagram: Let the side of the square be $x$, then
$$
tan beta = (x/2) / x = 1/2
$$
However
$$
tan(pi/8) = sqrt2-1 < 1/2
$$
So $beta ne pi/8$ and $a ne beta$.
Appendix:
The addition theorem for the tangent yields
$$
tan(2x) = frac2tan(x)1-tan^2(x)
$$
So
$$
underbracetan(2cdot pi/8)_1 = frac2tan(pi/8)1-tan^2(pi/8)
$$
and we have the quadratic equation in $u = tan(pi/8)$:
$$
beginalign
1-u^2 &= 2 u iff \
1 &= u^2 + 2u iff \
2 &= u^2 + 2u + 1 = (u+1)^2 iff \
u &= pmsqrt2-1
endalign
$$
where we need the positive solution.
answered Jul 21 at 16:51
mvw
30.4k22250
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up vote
2
down vote
From the diagram we have
$$
alpha + beta = pi/4
$$
If $alpha = beta$ this would give $alpha = beta = pi / 8$.
Again from the diagram: Let the side of the square be $x$, then
$$
tan beta = (x/2) / x = 1/2
$$
However
$$
tan(pi/8) = sqrt2-1 < 1/2
$$
So $beta ne pi/8$ and $a ne beta$.
Appendix:
The addition theorem for the tangent yields
$$
tan(2x) = frac2tan(x)1-tan^2(x)
$$
So
$$
underbracetan(2cdot pi/8)_1 = frac2tan(pi/8)1-tan^2(pi/8)
$$
and we have the quadratic equation in $u = tan(pi/8)$:
$$
beginalign
1-u^2 &= 2 u iff \
1 &= u^2 + 2u iff \
2 &= u^2 + 2u + 1 = (u+1)^2 iff \
u &= pmsqrt2-1
endalign
$$
where we need the positive solution.
answered Jul 21 at 16:51
mvw
30.4k22250
add a comment |Â
up vote
2
down vote
up vote
2
down vote
From the diagram we have
$$
alpha + beta = pi/4
$$
If $alpha = beta$ this would give $alpha = beta = pi / 8$.
Again from the diagram: Let the side of the square be $x$, then
$$
tan beta = (x/2) / x = 1/2
$$
However
$$
tan(pi/8) = sqrt2-1 < 1/2
$$
So $beta ne pi/8$ and $a ne beta$.
Appendix:
The addition theorem for the tangent yields
$$
tan(2x) = frac2tan(x)1-tan^2(x)
$$
So
$$
underbracetan(2cdot pi/8)_1 = frac2tan(pi/8)1-tan^2(pi/8)
$$
and we have the quadratic equation in $u = tan(pi/8)$:
$$
beginalign
1-u^2 &= 2 u iff \
1 &= u^2 + 2u iff \
2 &= u^2 + 2u + 1 = (u+1)^2 iff \
u &= pmsqrt2-1
endalign
$$
where we need the positive solution.
answered Jul 21 at 16:51
mvw
30.4k22250
From the diagram we have
$$
alpha + beta = pi/4
$$
If $alpha = beta$ this would give $alpha = beta = pi / 8$.
Again from the diagram: Let the side of the square be $x$, then
$$
tan beta = (x/2) / x = 1/2
$$
However
$$
tan(pi/8) = sqrt2-1 < 1/2
$$
So $beta ne pi/8$ and $a ne beta$.
Appendix:
The addition theorem for the tangent yields
$$
tan(2x) = frac2tan(x)1-tan^2(x)
$$
So
$$
underbracetan(2cdot pi/8)_1 = frac2tan(pi/8)1-tan^2(pi/8)
$$
and we have the quadratic equation in $u = tan(pi/8)$:
$$
beginalign
1-u^2 &= 2 u iff \
1 &= u^2 + 2u iff \
2 &= u^2 + 2u + 1 = (u+1)^2 iff \
u &= pmsqrt2-1
endalign
$$
where we need the positive solution.
answered Jul 21 at 16:51
mvw
30.4k22250
edited Jul 21 at 17:36
answered Jul 21 at 16:51
mvw
30.4k22250
answered Jul 21 at 16:51
mvw
30.4k22250
answered Jul 21 at 16:51
mvw
30.4k22250
30.4k22250
add a comment |Â
add a comment |Â
up vote
2
down vote
Drop a perpendicular from $M$ to the line $AB,$ meeting $AB$ at $P.$
That is, $P$ is the point on line $AB$ closest to $M.$
Since $angle ABM$ is not a right angle, $P$ is not $B,$
and therefore $MP < BM.$ Since $BM = CM,$ then $MP < CM.$
If $alpha = beta$ then the two right triangles
$triangle AMC$ and $triangle AMP$ would be congruent,
since all three angles and one side ($AM$) would be congruent.
But since $MP < CM,$ the triangles are not congruent.
Hence $alpha neq beta.$
answered Jul 21 at 17:38
David K
48.2k340107
add a comment |Â
up vote
2
down vote
Drop a perpendicular from $M$ to the line $AB,$ meeting $AB$ at $P.$
That is, $P$ is the point on line $AB$ closest to $M.$
Since $angle ABM$ is not a right angle, $P$ is not $B,$
and therefore $MP < BM.$ Since $BM = CM,$ then $MP < CM.$
If $alpha = beta$ then the two right triangles
$triangle AMC$ and $triangle AMP$ would be congruent,
since all three angles and one side ($AM$) would be congruent.
But since $MP < CM,$ the triangles are not congruent.
Hence $alpha neq beta.$
answered Jul 21 at 17:38
David K
48.2k340107
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Drop a perpendicular from $M$ to the line $AB,$ meeting $AB$ at $P.$
That is, $P$ is the point on line $AB$ closest to $M.$
Since $angle ABM$ is not a right angle, $P$ is not $B,$
and therefore $MP < BM.$ Since $BM = CM,$ then $MP < CM.$
If $alpha = beta$ then the two right triangles
$triangle AMC$ and $triangle AMP$ would be congruent,
since all three angles and one side ($AM$) would be congruent.
But since $MP < CM,$ the triangles are not congruent.
Hence $alpha neq beta.$
answered Jul 21 at 17:38
David K
48.2k340107
Drop a perpendicular from $M$ to the line $AB,$ meeting $AB$ at $P.$
That is, $P$ is the point on line $AB$ closest to $M.$
Since $angle ABM$ is not a right angle, $P$ is not $B,$
and therefore $MP < BM.$ Since $BM = CM,$ then $MP < CM.$
If $alpha = beta$ then the two right triangles
$triangle AMC$ and $triangle AMP$ would be congruent,
since all three angles and one side ($AM$) would be congruent.
But since $MP < CM,$ the triangles are not congruent.
Hence $alpha neq beta.$
answered Jul 21 at 17:38
David K
48.2k340107
answered Jul 21 at 17:38
David K
48.2k340107
answered Jul 21 at 17:38
David K
48.2k340107
answered Jul 21 at 17:38
David K
48.2k340107
48.2k340107
add a comment |Â
add a comment |Â
up vote
1
down vote
By the law of sines,
$$sqrt2fracsin(alpha)MB=frac1MA=fracsin(beta)MC.$$ Since $MC=MB$, we conclude that $sqrt2sin(alpha)=sin(beta)$. Clearly neither $alpha$ nor $beta$ are integer multiples of $pi$, so $alphaneqbeta$.
answered Jul 21 at 20:47
Alex S
17.6k11959
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up vote
1
down vote
By the law of sines,
$$sqrt2fracsin(alpha)MB=frac1MA=fracsin(beta)MC.$$ Since $MC=MB$, we conclude that $sqrt2sin(alpha)=sin(beta)$. Clearly neither $alpha$ nor $beta$ are integer multiples of $pi$, so $alphaneqbeta$.
answered Jul 21 at 20:47
Alex S
17.6k11959
add a comment |Â
up vote
1
down vote
up vote
1
down vote
By the law of sines,
$$sqrt2fracsin(alpha)MB=frac1MA=fracsin(beta)MC.$$ Since $MC=MB$, we conclude that $sqrt2sin(alpha)=sin(beta)$. Clearly neither $alpha$ nor $beta$ are integer multiples of $pi$, so $alphaneqbeta$.
answered Jul 21 at 20:47
Alex S
17.6k11959
By the law of sines,
$$sqrt2fracsin(alpha)MB=frac1MA=fracsin(beta)MC.$$ Since $MC=MB$, we conclude that $sqrt2sin(alpha)=sin(beta)$. Clearly neither $alpha$ nor $beta$ are integer multiples of $pi$, so $alphaneqbeta$.
answered Jul 21 at 20:47
Alex S
17.6k11959
answered Jul 21 at 20:47
Alex S
17.6k11959
answered Jul 21 at 20:47
Alex S
17.6k11959
answered Jul 21 at 20:47
Alex S
17.6k11959
17.6k11959
add a comment |Â
add a comment |Â
up vote
1
down vote
First assume that $alpha$ = $beta$ and that both would be equal to $fracpi8^circ$ (since $alpha = beta$ and $alpha + beta = fracpi4 to 2beta = fracpi4 to beta = fracpi8$).
If we call side $overlineAC$, n, then side $overlineMC$ is $fracn2$. Next, we solve for $overlineAM$ in terms of n to show that $alpha = beta$ for all values n.
$overlineAM^2 = n^2 + (fracn2)^2 = n^2 + fracn^24 = frac4n^2 + n^24 = frac5n^24$
$overlineAM = sqrtfrac5n^24 = fracsqrt5n^22 = fracnsqrt52$ by the pythagorean theorem
Now that we have $overlineAM$, we can find $cosbeta$.
$cosbeta=fracnoverlineAM = fracnfracnsqrt52 = frac2nnsqrt5 = frac2sqrt5 = frac2sqrt55$
$cosbeta = frac2sqrt55$
We have that $cosbeta = frac2sqrt55$ and we already established that $beta = fracpi8$.
This would mean that $cosfracpi8 = frac2sqrt55$. We can find $cosfracpi8 = fracsqrt2 + sqrt22$, which implies that $fracsqrt2 + sqrt22 = frac2sqrt55$.
If we square both sides, we get that $frac2+sqrt24 = frac2025 = frac45$. Then we can split the fraction, simplify, then subtract and simplify.
$frac24 + fracsqrt24 = frac45$
$frac12 + fracsqrt24 = frac45$
$fracsqrt24 = frac45 - frac12 $
$fracsqrt24 = frac310$
Next we can multiply both sides by 4 and conclude that...
$sqrt2 = frac1210 = frac65$
This is impossible since it has been proven that $sqrt2$ is irrational and can't be written as a ratio of 2 integers. This would mean that $sqrt2 neq frac65$ or $fracsqrt2 + sqrt22 neq frac2sqrt55$ or $cosfracpi8neq frac2sqrt55$
This means that $beta neq fracpi8$. Since $alpha + beta = fracpi4$, $beta$ would have to be equal to $fracpi8$, but we proved that it cant be. This means that $alpha neq beta$ since $beta$ can no longer be multiplied by 2 to be $fracpi4$
This method works with tangent and sine as well (I believe it's a lot less work if I had used tangent instead of cosine, but this was how I first solved it)
I'm a high-school student and this is my first response, feedback is always appreciated!
answered Jul 22 at 15:18
Jamar Sullivan
112
add a comment |Â
up vote
1
down vote
First assume that $alpha$ = $beta$ and that both would be equal to $fracpi8^circ$ (since $alpha = beta$ and $alpha + beta = fracpi4 to 2beta = fracpi4 to beta = fracpi8$).
If we call side $overlineAC$, n, then side $overlineMC$ is $fracn2$. Next, we solve for $overlineAM$ in terms of n to show that $alpha = beta$ for all values n.
$overlineAM^2 = n^2 + (fracn2)^2 = n^2 + fracn^24 = frac4n^2 + n^24 = frac5n^24$
$overlineAM = sqrtfrac5n^24 = fracsqrt5n^22 = fracnsqrt52$ by the pythagorean theorem
Now that we have $overlineAM$, we can find $cosbeta$.
$cosbeta=fracnoverlineAM = fracnfracnsqrt52 = frac2nnsqrt5 = frac2sqrt5 = frac2sqrt55$
$cosbeta = frac2sqrt55$
We have that $cosbeta = frac2sqrt55$ and we already established that $beta = fracpi8$.
This would mean that $cosfracpi8 = frac2sqrt55$. We can find $cosfracpi8 = fracsqrt2 + sqrt22$, which implies that $fracsqrt2 + sqrt22 = frac2sqrt55$.
If we square both sides, we get that $frac2+sqrt24 = frac2025 = frac45$. Then we can split the fraction, simplify, then subtract and simplify.
$frac24 + fracsqrt24 = frac45$
$frac12 + fracsqrt24 = frac45$
$fracsqrt24 = frac45 - frac12 $
$fracsqrt24 = frac310$
Next we can multiply both sides by 4 and conclude that...
$sqrt2 = frac1210 = frac65$
This is impossible since it has been proven that $sqrt2$ is irrational and can't be written as a ratio of 2 integers. This would mean that $sqrt2 neq frac65$ or $fracsqrt2 + sqrt22 neq frac2sqrt55$ or $cosfracpi8neq frac2sqrt55$
This means that $beta neq fracpi8$. Since $alpha + beta = fracpi4$, $beta$ would have to be equal to $fracpi8$, but we proved that it cant be. This means that $alpha neq beta$ since $beta$ can no longer be multiplied by 2 to be $fracpi4$
This method works with tangent and sine as well (I believe it's a lot less work if I had used tangent instead of cosine, but this was how I first solved it)
I'm a high-school student and this is my first response, feedback is always appreciated!
answered Jul 22 at 15:18
Jamar Sullivan
112
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First assume that $alpha$ = $beta$ and that both would be equal to $fracpi8^circ$ (since $alpha = beta$ and $alpha + beta = fracpi4 to 2beta = fracpi4 to beta = fracpi8$).
If we call side $overlineAC$, n, then side $overlineMC$ is $fracn2$. Next, we solve for $overlineAM$ in terms of n to show that $alpha = beta$ for all values n.
$overlineAM^2 = n^2 + (fracn2)^2 = n^2 + fracn^24 = frac4n^2 + n^24 = frac5n^24$
$overlineAM = sqrtfrac5n^24 = fracsqrt5n^22 = fracnsqrt52$ by the pythagorean theorem
Now that we have $overlineAM$, we can find $cosbeta$.
$cosbeta=fracnoverlineAM = fracnfracnsqrt52 = frac2nnsqrt5 = frac2sqrt5 = frac2sqrt55$
$cosbeta = frac2sqrt55$
We have that $cosbeta = frac2sqrt55$ and we already established that $beta = fracpi8$.
This would mean that $cosfracpi8 = frac2sqrt55$. We can find $cosfracpi8 = fracsqrt2 + sqrt22$, which implies that $fracsqrt2 + sqrt22 = frac2sqrt55$.
If we square both sides, we get that $frac2+sqrt24 = frac2025 = frac45$. Then we can split the fraction, simplify, then subtract and simplify.
$frac24 + fracsqrt24 = frac45$
$frac12 + fracsqrt24 = frac45$
$fracsqrt24 = frac45 - frac12 $
$fracsqrt24 = frac310$
Next we can multiply both sides by 4 and conclude that...
$sqrt2 = frac1210 = frac65$
This is impossible since it has been proven that $sqrt2$ is irrational and can't be written as a ratio of 2 integers. This would mean that $sqrt2 neq frac65$ or $fracsqrt2 + sqrt22 neq frac2sqrt55$ or $cosfracpi8neq frac2sqrt55$
This means that $beta neq fracpi8$. Since $alpha + beta = fracpi4$, $beta$ would have to be equal to $fracpi8$, but we proved that it cant be. This means that $alpha neq beta$ since $beta$ can no longer be multiplied by 2 to be $fracpi4$
This method works with tangent and sine as well (I believe it's a lot less work if I had used tangent instead of cosine, but this was how I first solved it)
I'm a high-school student and this is my first response, feedback is always appreciated!
answered Jul 22 at 15:18
Jamar Sullivan
112
First assume that $alpha$ = $beta$ and that both would be equal to $fracpi8^circ$ (since $alpha = beta$ and $alpha + beta = fracpi4 to 2beta = fracpi4 to beta = fracpi8$).
If we call side $overlineAC$, n, then side $overlineMC$ is $fracn2$. Next, we solve for $overlineAM$ in terms of n to show that $alpha = beta$ for all values n.
$overlineAM^2 = n^2 + (fracn2)^2 = n^2 + fracn^24 = frac4n^2 + n^24 = frac5n^24$
$overlineAM = sqrtfrac5n^24 = fracsqrt5n^22 = fracnsqrt52$ by the pythagorean theorem
Now that we have $overlineAM$, we can find $cosbeta$.
$cosbeta=fracnoverlineAM = fracnfracnsqrt52 = frac2nnsqrt5 = frac2sqrt5 = frac2sqrt55$
$cosbeta = frac2sqrt55$
We have that $cosbeta = frac2sqrt55$ and we already established that $beta = fracpi8$.
This would mean that $cosfracpi8 = frac2sqrt55$. We can find $cosfracpi8 = fracsqrt2 + sqrt22$, which implies that $fracsqrt2 + sqrt22 = frac2sqrt55$.
If we square both sides, we get that $frac2+sqrt24 = frac2025 = frac45$. Then we can split the fraction, simplify, then subtract and simplify.
$frac24 + fracsqrt24 = frac45$
$frac12 + fracsqrt24 = frac45$
$fracsqrt24 = frac45 - frac12 $
$fracsqrt24 = frac310$
Next we can multiply both sides by 4 and conclude that...
$sqrt2 = frac1210 = frac65$
This is impossible since it has been proven that $sqrt2$ is irrational and can't be written as a ratio of 2 integers. This would mean that $sqrt2 neq frac65$ or $fracsqrt2 + sqrt22 neq frac2sqrt55$ or $cosfracpi8neq frac2sqrt55$
This means that $beta neq fracpi8$. Since $alpha + beta = fracpi4$, $beta$ would have to be equal to $fracpi8$, but we proved that it cant be. This means that $alpha neq beta$ since $beta$ can no longer be multiplied by 2 to be $fracpi4$
This method works with tangent and sine as well (I believe it's a lot less work if I had used tangent instead of cosine, but this was how I first solved it)
I'm a high-school student and this is my first response, feedback is always appreciated!
answered Jul 22 at 15:18
Jamar Sullivan
112
answered Jul 22 at 15:18
Jamar Sullivan
112
answered Jul 22 at 15:18
Jamar Sullivan
112
answered Jul 22 at 15:18
Jamar Sullivan
112
112
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up vote
1
down vote
Here's one intuitive way to analyze the problem:
If you're standing on point $A$ and looking at a car driving at constant speed on a straight line $BC$, you'll have to turn your head very fast when the car is closest to you (on point $C$) and slow the head rotation as the car drives away.
answered Jul 22 at 20:41
Eric Duminil
1,6961516
add a comment |Â
up vote
1
down vote
Here's one intuitive way to analyze the problem:
If you're standing on point $A$ and looking at a car driving at constant speed on a straight line $BC$, you'll have to turn your head very fast when the car is closest to you (on point $C$) and slow the head rotation as the car drives away.
answered Jul 22 at 20:41
Eric Duminil
1,6961516
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here's one intuitive way to analyze the problem:
If you're standing on point $A$ and looking at a car driving at constant speed on a straight line $BC$, you'll have to turn your head very fast when the car is closest to you (on point $C$) and slow the head rotation as the car drives away.
answered Jul 22 at 20:41
Eric Duminil
1,6961516
Here's one intuitive way to analyze the problem:
If you're standing on point $A$ and looking at a car driving at constant speed on a straight line $BC$, you'll have to turn your head very fast when the car is closest to you (on point $C$) and slow the head rotation as the car drives away.
answered Jul 22 at 20:41
Eric Duminil
1,6961516
answered Jul 22 at 20:41
Eric Duminil
1,6961516
answered Jul 22 at 20:41
Eric Duminil
1,6961516
answered Jul 22 at 20:41
Eric Duminil
1,6961516
1,6961516
add a comment |Â
add a comment |Â
up vote
0
down vote
The figure $triangle ABC$ is the half of a square therefore the sides $overlineAB$ and $overlineBC$ are of equal length. Now consider the triangle $triangle ABM$. It is a rectangular triangle such as the original one.
Therefore the tangens of the angle $alpha$ is given by
$$tan(alpha)~=~fracoverlineBMoverlineAB$$
hence $M$ is the middle of the side $overlineBC$ this simplifies to
$$tan(alpha)~=~fracoverlineBMoverlineAB~=~fracfracoverlineBC2overlineAB~=~fracfracoverlineBC2overlineBC~=~frac12\
alpha~=~arctanleft(frac12right)$$
Now go back to the triangle $triangle ABC$. The tangens of the angle $alpha + beta$ is given by
$$tan(alpha + beta)~=~fracoverlineBCoverlineAB$$
which equals $1$. The tangens of the sum of two different angles is given by
$$tan(x+y)~=~fractan(x)+tan(y)1-tan(x)tan(y)$$
with $x=alpha$,$y=beta$ and $tan(alpha)=frac12$ you get
$$beginalign
tan(alpha+beta)~=~1~=~fracfrac12+tan(beta)1-frac12tan(beta)~iff~1-frac12tan(beta)~&=~frac12+tan(beta)\
tan(beta)&=frac13\
beta&=arctanleft(frac13right)
endalign$$
Hence $arctanleft(frac12right)neqarctanleft(frac13right)$ and so $alphaneqbeta$.
answered Jul 21 at 17:13
mrtaurho
700219
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The figure $triangle ABC$ is the half of a square therefore the sides $overlineAB$ and $overlineBC$ are of equal length. Now consider the triangle $triangle ABM$. It is a rectangular triangle such as the original one.
Therefore the tangens of the angle $alpha$ is given by
$$tan(alpha)~=~fracoverlineBMoverlineAB$$
hence $M$ is the middle of the side $overlineBC$ this simplifies to
$$tan(alpha)~=~fracoverlineBMoverlineAB~=~fracfracoverlineBC2overlineAB~=~fracfracoverlineBC2overlineBC~=~frac12\
alpha~=~arctanleft(frac12right)$$
Now go back to the triangle $triangle ABC$. The tangens of the angle $alpha + beta$ is given by
$$tan(alpha + beta)~=~fracoverlineBCoverlineAB$$
which equals $1$. The tangens of the sum of two different angles is given by
$$tan(x+y)~=~fractan(x)+tan(y)1-tan(x)tan(y)$$
with $x=alpha$,$y=beta$ and $tan(alpha)=frac12$ you get
$$beginalign
tan(alpha+beta)~=~1~=~fracfrac12+tan(beta)1-frac12tan(beta)~iff~1-frac12tan(beta)~&=~frac12+tan(beta)\
tan(beta)&=frac13\
beta&=arctanleft(frac13right)
endalign$$
Hence $arctanleft(frac12right)neqarctanleft(frac13right)$ and so $alphaneqbeta$.
answered Jul 21 at 17:13
mrtaurho
700219
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The figure $triangle ABC$ is the half of a square therefore the sides $overlineAB$ and $overlineBC$ are of equal length. Now consider the triangle $triangle ABM$. It is a rectangular triangle such as the original one.
Therefore the tangens of the angle $alpha$ is given by
$$tan(alpha)~=~fracoverlineBMoverlineAB$$
hence $M$ is the middle of the side $overlineBC$ this simplifies to
$$tan(alpha)~=~fracoverlineBMoverlineAB~=~fracfracoverlineBC2overlineAB~=~fracfracoverlineBC2overlineBC~=~frac12\
alpha~=~arctanleft(frac12right)$$
Now go back to the triangle $triangle ABC$. The tangens of the angle $alpha + beta$ is given by
$$tan(alpha + beta)~=~fracoverlineBCoverlineAB$$
which equals $1$. The tangens of the sum of two different angles is given by
$$tan(x+y)~=~fractan(x)+tan(y)1-tan(x)tan(y)$$
with $x=alpha$,$y=beta$ and $tan(alpha)=frac12$ you get
$$beginalign
tan(alpha+beta)~=~1~=~fracfrac12+tan(beta)1-frac12tan(beta)~iff~1-frac12tan(beta)~&=~frac12+tan(beta)\
tan(beta)&=frac13\
beta&=arctanleft(frac13right)
endalign$$
Hence $arctanleft(frac12right)neqarctanleft(frac13right)$ and so $alphaneqbeta$.
answered Jul 21 at 17:13
mrtaurho
700219
The figure $triangle ABC$ is the half of a square therefore the sides $overlineAB$ and $overlineBC$ are of equal length. Now consider the triangle $triangle ABM$. It is a rectangular triangle such as the original one.
Therefore the tangens of the angle $alpha$ is given by
$$tan(alpha)~=~fracoverlineBMoverlineAB$$
hence $M$ is the middle of the side $overlineBC$ this simplifies to
$$tan(alpha)~=~fracoverlineBMoverlineAB~=~fracfracoverlineBC2overlineAB~=~fracfracoverlineBC2overlineBC~=~frac12\
alpha~=~arctanleft(frac12right)$$
Now go back to the triangle $triangle ABC$. The tangens of the angle $alpha + beta$ is given by
$$tan(alpha + beta)~=~fracoverlineBCoverlineAB$$
which equals $1$. The tangens of the sum of two different angles is given by
$$tan(x+y)~=~fractan(x)+tan(y)1-tan(x)tan(y)$$
with $x=alpha$,$y=beta$ and $tan(alpha)=frac12$ you get
$$beginalign
tan(alpha+beta)~=~1~=~fracfrac12+tan(beta)1-frac12tan(beta)~iff~1-frac12tan(beta)~&=~frac12+tan(beta)\
tan(beta)&=frac13\
beta&=arctanleft(frac13right)
endalign$$
Hence $arctanleft(frac12right)neqarctanleft(frac13right)$ and so $alphaneqbeta$.
answered Jul 21 at 17:13
mrtaurho
700219
edited Jul 21 at 17:31
answered Jul 21 at 17:13
mrtaurho
700219
answered Jul 21 at 17:13
mrtaurho
700219
answered Jul 21 at 17:13
mrtaurho
700219
700219
add a comment |Â
add a comment |Â
up vote
0
down vote
Assume $alpha$ and $beta$ are both equal to $22.5^circ$ (sum to forty-five).
The angle $angle ABM$ must equal $45^circ$ because it bisects the square.
The angle $angle ACM$ occupies a square corner, so it must equal $90^circ$.
$alpha + angle BMA$ must not sum to more than $135^circ$ ($180^circ$ $-$ $45^circ$ from $angle ABM$).
If alpha is equal to beta and both equal $22.5^circ$, then BMA
must equal $112.5$ degrees ($135^circ - 22.5^circ$).
BMA + AMC must equal less than $180$ degrees (one hemicircle); if $angle BMA = 112.5^circ$, then $angle AMC$ must be $67.5^circ$.
Given $beta = alpha = 22.5^circ$, $angle AMC + angle ACM$ must equal $157.5^circ$ ($180$ minus $beta$); given the same assumption, $angle AMC$ is equal to $67.5^circ$ and sums with $angle ACM$ to $147.5^circ$.
Alpha must not be equal to beta.
answered Jul 23 at 3:16
Paul Ferris
33
add a comment |Â
up vote
0
down vote
Assume $alpha$ and $beta$ are both equal to $22.5^circ$ (sum to forty-five).
The angle $angle ABM$ must equal $45^circ$ because it bisects the square.
The angle $angle ACM$ occupies a square corner, so it must equal $90^circ$.
$alpha + angle BMA$ must not sum to more than $135^circ$ ($180^circ$ $-$ $45^circ$ from $angle ABM$).
If alpha is equal to beta and both equal $22.5^circ$, then BMA
must equal $112.5$ degrees ($135^circ - 22.5^circ$).
BMA + AMC must equal less than $180$ degrees (one hemicircle); if $angle BMA = 112.5^circ$, then $angle AMC$ must be $67.5^circ$.
Given $beta = alpha = 22.5^circ$, $angle AMC + angle ACM$ must equal $157.5^circ$ ($180$ minus $beta$); given the same assumption, $angle AMC$ is equal to $67.5^circ$ and sums with $angle ACM$ to $147.5^circ$.
Alpha must not be equal to beta.
answered Jul 23 at 3:16
Paul Ferris
33
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assume $alpha$ and $beta$ are both equal to $22.5^circ$ (sum to forty-five).
The angle $angle ABM$ must equal $45^circ$ because it bisects the square.
The angle $angle ACM$ occupies a square corner, so it must equal $90^circ$.
$alpha + angle BMA$ must not sum to more than $135^circ$ ($180^circ$ $-$ $45^circ$ from $angle ABM$).
If alpha is equal to beta and both equal $22.5^circ$, then BMA
must equal $112.5$ degrees ($135^circ - 22.5^circ$).
BMA + AMC must equal less than $180$ degrees (one hemicircle); if $angle BMA = 112.5^circ$, then $angle AMC$ must be $67.5^circ$.
Given $beta = alpha = 22.5^circ$, $angle AMC + angle ACM$ must equal $157.5^circ$ ($180$ minus $beta$); given the same assumption, $angle AMC$ is equal to $67.5^circ$ and sums with $angle ACM$ to $147.5^circ$.
Alpha must not be equal to beta.
answered Jul 23 at 3:16
Paul Ferris
33
Assume $alpha$ and $beta$ are both equal to $22.5^circ$ (sum to forty-five).
The angle $angle ABM$ must equal $45^circ$ because it bisects the square.
The angle $angle ACM$ occupies a square corner, so it must equal $90^circ$.
$alpha + angle BMA$ must not sum to more than $135^circ$ ($180^circ$ $-$ $45^circ$ from $angle ABM$).
If alpha is equal to beta and both equal $22.5^circ$, then BMA
must equal $112.5$ degrees ($135^circ - 22.5^circ$).
BMA + AMC must equal less than $180$ degrees (one hemicircle); if $angle BMA = 112.5^circ$, then $angle AMC$ must be $67.5^circ$.
Given $beta = alpha = 22.5^circ$, $angle AMC + angle ACM$ must equal $157.5^circ$ ($180$ minus $beta$); given the same assumption, $angle AMC$ is equal to $67.5^circ$ and sums with $angle ACM$ to $147.5^circ$.
Alpha must not be equal to beta.
answered Jul 23 at 3:16
Paul Ferris
33
answered Jul 23 at 3:16
Paul Ferris
33
answered Jul 23 at 3:16
Paul Ferris
33
answered Jul 23 at 3:16
Paul Ferris
33
33
add a comment |Â
add a comment |Â
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8
This is an illustration that the tangent function is nonlinear.
– Yves Daoust
Jul 21 at 17:03
1
Fun question! Thanks for sharing :)
– Sambo
Jul 21 at 17:09