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Additional solutions and generalizations of an 8th grade Olympiad problem about a cube
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This year, I was the author of the following problem which came at the county level of the Math Olympiad at 8th grade:
Consider cube $ABCDA^prime B^prime C^prime D^prime$ with faces $square ABCD$ and $square A^prime B^prime C^prime D^prime$ connected by edges $[AA^prime]$, $[BB^prime]$, $[CC^prime]$, $[DD^prime]$. Let $M$ be the middle of the side $[AB]$, and $E$ be the symmetric of $D^prime$ with respect to $M$. Prove that $DB^primeperp(A^prime C^prime E)$.
The interesting thing was that this was regarded as difficult, even by teachers I talked to about this problem. I will leave to you to find the solutions to this (there is a very simple one:
There is actually only one plane perpendicular on $DB^prime$ which contains the line $A^prime C^prime$, i.e. $A^prime C^prime B$, so we only have to prove that this plane contains $E$, which can be done by looking at the triangle $D^prime C^prime E$.
another one would be:
This time, let's prove by seeingthe $DB^primeperp A^prime C^prime$. Also, we will prove that $DB^primeperp A^prime E$. Let $F$ the symmetric of $C$ with respect to $D$. The idea is to calculate each of the sides of the triangle $A^prime FE$ and prove that it is right angled, thus we will get the conclusion.
I'm really interested in how many new ones we can find).
Also, I am really curious on this particular configuration which can lead to many questions (as an example: if we denote by $O$ the center of the face $A^prime B^prime C^prime D^prime$, we can ask many questions about the trapezoid determined by the intersection of the plane $(OAE)$. Can you figure out some?). Even further, I think it would be very interesting to generalize the original problem in $4$ dimensions.
I really hope that this can lead to many problems for us to solve.
geometry euclidean-geometry 3d
edited Jul 22 at 20:48
greedoid
26.1k93473
asked Jul 22 at 11:42
Andrei Cataron
1178
add a comment |Â
up vote
1
down vote
favorite
This year, I was the author of the following problem which came at the county level of the Math Olympiad at 8th grade:
Consider cube $ABCDA^prime B^prime C^prime D^prime$ with faces $square ABCD$ and $square A^prime B^prime C^prime D^prime$ connected by edges $[AA^prime]$, $[BB^prime]$, $[CC^prime]$, $[DD^prime]$. Let $M$ be the middle of the side $[AB]$, and $E$ be the symmetric of $D^prime$ with respect to $M$. Prove that $DB^primeperp(A^prime C^prime E)$.
The interesting thing was that this was regarded as difficult, even by teachers I talked to about this problem. I will leave to you to find the solutions to this (there is a very simple one:
There is actually only one plane perpendicular on $DB^prime$ which contains the line $A^prime C^prime$, i.e. $A^prime C^prime B$, so we only have to prove that this plane contains $E$, which can be done by looking at the triangle $D^prime C^prime E$.
another one would be:
This time, let's prove by seeingthe $DB^primeperp A^prime C^prime$. Also, we will prove that $DB^primeperp A^prime E$. Let $F$ the symmetric of $C$ with respect to $D$. The idea is to calculate each of the sides of the triangle $A^prime FE$ and prove that it is right angled, thus we will get the conclusion.
I'm really interested in how many new ones we can find).
Also, I am really curious on this particular configuration which can lead to many questions (as an example: if we denote by $O$ the center of the face $A^prime B^prime C^prime D^prime$, we can ask many questions about the trapezoid determined by the intersection of the plane $(OAE)$. Can you figure out some?). Even further, I think it would be very interesting to generalize the original problem in $4$ dimensions.
I really hope that this can lead to many problems for us to solve.
geometry euclidean-geometry 3d
edited Jul 22 at 20:48
greedoid
26.1k93473
asked Jul 22 at 11:42
Andrei Cataron
1178
You seem to be asking two questions: What are other solutions to the problem? and What are generalizations of the problem? These should be posted separately. (You could/should provide links from one to the other to provide context.) BTW: If you want "new" solutions, then you should post the solutions you know so that answerers don't duplicate the effort.
– Blue
Jul 22 at 13:11
goo.gl/images/1iAyQB here is an image of the cube
– Andrei Cataron
Jul 22 at 13:16
About the solutions I have, I'm afraid that if one sees them, it may limit the vision towards others. What do you think?
– Andrei Cataron
Jul 22 at 13:18
1
Math.SE is a question-and-answer site, intended for asking about things you don't know. Therefore, you should post what you know. (You can "hide" solutions by typing>!in front of the text.) It's no fun for answerers when they devise a clever solution, and go through the trouble of posting it with properly-formatted mathematical expressions and fancy diagrams, only to have the answerer say, "Oh, I already know that one." That's just a waste of everyone's time.
– Blue
Jul 22 at 13:36
1
I will edit them in right away.
– Andrei Cataron
Jul 22 at 14:01
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This year, I was the author of the following problem which came at the county level of the Math Olympiad at 8th grade:
Consider cube $ABCDA^prime B^prime C^prime D^prime$ with faces $square ABCD$ and $square A^prime B^prime C^prime D^prime$ connected by edges $[AA^prime]$, $[BB^prime]$, $[CC^prime]$, $[DD^prime]$. Let $M$ be the middle of the side $[AB]$, and $E$ be the symmetric of $D^prime$ with respect to $M$. Prove that $DB^primeperp(A^prime C^prime E)$.
The interesting thing was that this was regarded as difficult, even by teachers I talked to about this problem. I will leave to you to find the solutions to this (there is a very simple one:
There is actually only one plane perpendicular on $DB^prime$ which contains the line $A^prime C^prime$, i.e. $A^prime C^prime B$, so we only have to prove that this plane contains $E$, which can be done by looking at the triangle $D^prime C^prime E$.
another one would be:
This time, let's prove by seeingthe $DB^primeperp A^prime C^prime$. Also, we will prove that $DB^primeperp A^prime E$. Let $F$ the symmetric of $C$ with respect to $D$. The idea is to calculate each of the sides of the triangle $A^prime FE$ and prove that it is right angled, thus we will get the conclusion.
I'm really interested in how many new ones we can find).
Also, I am really curious on this particular configuration which can lead to many questions (as an example: if we denote by $O$ the center of the face $A^prime B^prime C^prime D^prime$, we can ask many questions about the trapezoid determined by the intersection of the plane $(OAE)$. Can you figure out some?). Even further, I think it would be very interesting to generalize the original problem in $4$ dimensions.
I really hope that this can lead to many problems for us to solve.
geometry euclidean-geometry 3d
edited Jul 22 at 20:48
greedoid
26.1k93473
asked Jul 22 at 11:42
Andrei Cataron
1178
This year, I was the author of the following problem which came at the county level of the Math Olympiad at 8th grade:
Consider cube $ABCDA^prime B^prime C^prime D^prime$ with faces $square ABCD$ and $square A^prime B^prime C^prime D^prime$ connected by edges $[AA^prime]$, $[BB^prime]$, $[CC^prime]$, $[DD^prime]$. Let $M$ be the middle of the side $[AB]$, and $E$ be the symmetric of $D^prime$ with respect to $M$. Prove that $DB^primeperp(A^prime C^prime E)$.
The interesting thing was that this was regarded as difficult, even by teachers I talked to about this problem. I will leave to you to find the solutions to this (there is a very simple one:
There is actually only one plane perpendicular on $DB^prime$ which contains the line $A^prime C^prime$, i.e. $A^prime C^prime B$, so we only have to prove that this plane contains $E$, which can be done by looking at the triangle $D^prime C^prime E$.
another one would be:
This time, let's prove by seeingthe $DB^primeperp A^prime C^prime$. Also, we will prove that $DB^primeperp A^prime E$. Let $F$ the symmetric of $C$ with respect to $D$. The idea is to calculate each of the sides of the triangle $A^prime FE$ and prove that it is right angled, thus we will get the conclusion.
I'm really interested in how many new ones we can find).
Also, I am really curious on this particular configuration which can lead to many questions (as an example: if we denote by $O$ the center of the face $A^prime B^prime C^prime D^prime$, we can ask many questions about the trapezoid determined by the intersection of the plane $(OAE)$. Can you figure out some?). Even further, I think it would be very interesting to generalize the original problem in $4$ dimensions.
I really hope that this can lead to many problems for us to solve.
geometry euclidean-geometry 3d
edited Jul 22 at 20:48
greedoid
26.1k93473
asked Jul 22 at 11:42
Andrei Cataron
1178
edited Jul 22 at 20:48
greedoid
26.1k93473
edited Jul 22 at 20:48
greedoid
26.1k93473
edited Jul 22 at 20:48
greedoid
26.1k93473
26.1k93473
asked Jul 22 at 11:42
Andrei Cataron
1178
asked Jul 22 at 11:42
Andrei Cataron
1178
asked Jul 22 at 11:42
Andrei Cataron
1178
1178
You seem to be asking two questions: What are other solutions to the problem? and What are generalizations of the problem? These should be posted separately. (You could/should provide links from one to the other to provide context.) BTW: If you want "new" solutions, then you should post the solutions you know so that answerers don't duplicate the effort.
– Blue
Jul 22 at 13:11
goo.gl/images/1iAyQB here is an image of the cube
– Andrei Cataron
Jul 22 at 13:16
About the solutions I have, I'm afraid that if one sees them, it may limit the vision towards others. What do you think?
– Andrei Cataron
Jul 22 at 13:18
1
Math.SE is a question-and-answer site, intended for asking about things you don't know. Therefore, you should post what you know. (You can "hide" solutions by typing>!in front of the text.) It's no fun for answerers when they devise a clever solution, and go through the trouble of posting it with properly-formatted mathematical expressions and fancy diagrams, only to have the answerer say, "Oh, I already know that one." That's just a waste of everyone's time.
– Blue
Jul 22 at 13:36
1
I will edit them in right away.
– Andrei Cataron
Jul 22 at 14:01
add a comment |Â
You seem to be asking two questions: What are other solutions to the problem? and What are generalizations of the problem? These should be posted separately. (You could/should provide links from one to the other to provide context.) BTW: If you want "new" solutions, then you should post the solutions you know so that answerers don't duplicate the effort.
– Blue
Jul 22 at 13:11
goo.gl/images/1iAyQB here is an image of the cube
– Andrei Cataron
Jul 22 at 13:16
About the solutions I have, I'm afraid that if one sees them, it may limit the vision towards others. What do you think?
– Andrei Cataron
Jul 22 at 13:18
1
Math.SE is a question-and-answer site, intended for asking about things you don't know. Therefore, you should post what you know. (You can "hide" solutions by typing>!in front of the text.) It's no fun for answerers when they devise a clever solution, and go through the trouble of posting it with properly-formatted mathematical expressions and fancy diagrams, only to have the answerer say, "Oh, I already know that one." That's just a waste of everyone's time.
– Blue
Jul 22 at 13:36
1
I will edit them in right away.
– Andrei Cataron
Jul 22 at 14:01
You seem to be asking two questions: What are other solutions to the problem? and What are generalizations of the problem? These should be posted separately. (You could/should provide links from one to the other to provide context.) BTW: If you want "new" solutions, then you should post the solutions you know so that answerers don't duplicate the effort.
– Blue
Jul 22 at 13:11
You seem to be asking two questions: What are other solutions to the problem? and What are generalizations of the problem? These should be posted separately. (You could/should provide links from one to the other to provide context.) BTW: If you want "new" solutions, then you should post the solutions you know so that answerers don't duplicate the effort.
– Blue
Jul 22 at 13:11
goo.gl/images/1iAyQB here is an image of the cube
– Andrei Cataron
Jul 22 at 13:16
goo.gl/images/1iAyQB here is an image of the cube
– Andrei Cataron
Jul 22 at 13:16
About the solutions I have, I'm afraid that if one sees them, it may limit the vision towards others. What do you think?
– Andrei Cataron
Jul 22 at 13:18
About the solutions I have, I'm afraid that if one sees them, it may limit the vision towards others. What do you think?
– Andrei Cataron
Jul 22 at 13:18
1
1
Math.SE is a question-and-answer site, intended for asking about things you don't know. Therefore, you should post what you know. (You can "hide" solutions by typing
>! in front of the text.) It's no fun for answerers when they devise a clever solution, and go through the trouble of posting it with properly-formatted mathematical expressions and fancy diagrams, only to have the answerer say, "Oh, I already know that one." That's just a waste of everyone's time.– Blue
Jul 22 at 13:36
Math.SE is a question-and-answer site, intended for asking about things you don't know. Therefore, you should post what you know. (You can "hide" solutions by typing
>! in front of the text.) It's no fun for answerers when they devise a clever solution, and go through the trouble of posting it with properly-formatted mathematical expressions and fancy diagrams, only to have the answerer say, "Oh, I already know that one." That's just a waste of everyone's time.– Blue
Jul 22 at 13:36
1
1
I will edit them in right away.
– Andrei Cataron
Jul 22 at 14:01
I will edit them in right away.
– Andrei Cataron
Jul 22 at 14:01
add a comment |Â
2 Answers
2
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oldest
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up vote
2
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Perhaps I'm being pessimistic, but this problem seems to simply give free points to those who know basic vector operations.
Let $A (0,0,0), B (1,0,0), C (1,1,0), D (0,1,0)$ and $X'=X+(0,0,1).$
It is easy to verify that $DB'= (1,-1,1), M = (0.5,0,0), E (1,-1,-1)$
Let $V = A'E times A'C' = (1,-1,-2) times (1,1,0) = (2,-2,2)$ be a vector normal to $(A'C'E).$ We have $V = 2 cdot DB',$ from which the result follows.
answered Jul 22 at 14:27
Display name
623211
add a comment |Â
up vote
1
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It is enough to prove $DB'bot AC'$ and $DB'bot AE$. Say $AB =a$
We will use this property:
A line $XY$ is perpendicular to a line $AB$ iff $$AX^2-AY^2 =BX^2-BY^2$$
$bullet DB'bot AC':$ Since $A'D = DC'$ and $A'B' = C'B'$ we have $$DC'^2-B'C'^2 = DA'^2-B'A'^2 $$
$bullet DB'bot AE:$ Since $AB' = a$, $A'D =asqrt2$, $EB' = asqrt5$ and $DE = asqrt6$ we have $$DE^2-B'E^2 = 6a^2- 5a^2 = 2a^2-a^2 =DA'^2-B'A'^2 $$
and we are done.
answered Jul 22 at 15:47
greedoid
26.1k93473
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Perhaps I'm being pessimistic, but this problem seems to simply give free points to those who know basic vector operations.
Let $A (0,0,0), B (1,0,0), C (1,1,0), D (0,1,0)$ and $X'=X+(0,0,1).$
It is easy to verify that $DB'= (1,-1,1), M = (0.5,0,0), E (1,-1,-1)$
Let $V = A'E times A'C' = (1,-1,-2) times (1,1,0) = (2,-2,2)$ be a vector normal to $(A'C'E).$ We have $V = 2 cdot DB',$ from which the result follows.
answered Jul 22 at 14:27
Display name
623211
add a comment |Â
up vote
2
down vote
Perhaps I'm being pessimistic, but this problem seems to simply give free points to those who know basic vector operations.
Let $A (0,0,0), B (1,0,0), C (1,1,0), D (0,1,0)$ and $X'=X+(0,0,1).$
It is easy to verify that $DB'= (1,-1,1), M = (0.5,0,0), E (1,-1,-1)$
Let $V = A'E times A'C' = (1,-1,-2) times (1,1,0) = (2,-2,2)$ be a vector normal to $(A'C'E).$ We have $V = 2 cdot DB',$ from which the result follows.
answered Jul 22 at 14:27
Display name
623211
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Perhaps I'm being pessimistic, but this problem seems to simply give free points to those who know basic vector operations.
Let $A (0,0,0), B (1,0,0), C (1,1,0), D (0,1,0)$ and $X'=X+(0,0,1).$
It is easy to verify that $DB'= (1,-1,1), M = (0.5,0,0), E (1,-1,-1)$
Let $V = A'E times A'C' = (1,-1,-2) times (1,1,0) = (2,-2,2)$ be a vector normal to $(A'C'E).$ We have $V = 2 cdot DB',$ from which the result follows.
answered Jul 22 at 14:27
Display name
623211
Perhaps I'm being pessimistic, but this problem seems to simply give free points to those who know basic vector operations.
Let $A (0,0,0), B (1,0,0), C (1,1,0), D (0,1,0)$ and $X'=X+(0,0,1).$
It is easy to verify that $DB'= (1,-1,1), M = (0.5,0,0), E (1,-1,-1)$
Let $V = A'E times A'C' = (1,-1,-2) times (1,1,0) = (2,-2,2)$ be a vector normal to $(A'C'E).$ We have $V = 2 cdot DB',$ from which the result follows.
answered Jul 22 at 14:27
Display name
623211
answered Jul 22 at 14:27
Display name
623211
answered Jul 22 at 14:27
Display name
623211
answered Jul 22 at 14:27
Display name
623211
623211
add a comment |Â
add a comment |Â
up vote
1
down vote
It is enough to prove $DB'bot AC'$ and $DB'bot AE$. Say $AB =a$
We will use this property:
A line $XY$ is perpendicular to a line $AB$ iff $$AX^2-AY^2 =BX^2-BY^2$$
$bullet DB'bot AC':$ Since $A'D = DC'$ and $A'B' = C'B'$ we have $$DC'^2-B'C'^2 = DA'^2-B'A'^2 $$
$bullet DB'bot AE:$ Since $AB' = a$, $A'D =asqrt2$, $EB' = asqrt5$ and $DE = asqrt6$ we have $$DE^2-B'E^2 = 6a^2- 5a^2 = 2a^2-a^2 =DA'^2-B'A'^2 $$
and we are done.
answered Jul 22 at 15:47
greedoid
26.1k93473
add a comment |Â
up vote
1
down vote
It is enough to prove $DB'bot AC'$ and $DB'bot AE$. Say $AB =a$
We will use this property:
A line $XY$ is perpendicular to a line $AB$ iff $$AX^2-AY^2 =BX^2-BY^2$$
$bullet DB'bot AC':$ Since $A'D = DC'$ and $A'B' = C'B'$ we have $$DC'^2-B'C'^2 = DA'^2-B'A'^2 $$
$bullet DB'bot AE:$ Since $AB' = a$, $A'D =asqrt2$, $EB' = asqrt5$ and $DE = asqrt6$ we have $$DE^2-B'E^2 = 6a^2- 5a^2 = 2a^2-a^2 =DA'^2-B'A'^2 $$
and we are done.
answered Jul 22 at 15:47
greedoid
26.1k93473
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is enough to prove $DB'bot AC'$ and $DB'bot AE$. Say $AB =a$
We will use this property:
A line $XY$ is perpendicular to a line $AB$ iff $$AX^2-AY^2 =BX^2-BY^2$$
$bullet DB'bot AC':$ Since $A'D = DC'$ and $A'B' = C'B'$ we have $$DC'^2-B'C'^2 = DA'^2-B'A'^2 $$
$bullet DB'bot AE:$ Since $AB' = a$, $A'D =asqrt2$, $EB' = asqrt5$ and $DE = asqrt6$ we have $$DE^2-B'E^2 = 6a^2- 5a^2 = 2a^2-a^2 =DA'^2-B'A'^2 $$
and we are done.
answered Jul 22 at 15:47
greedoid
26.1k93473
It is enough to prove $DB'bot AC'$ and $DB'bot AE$. Say $AB =a$
We will use this property:
A line $XY$ is perpendicular to a line $AB$ iff $$AX^2-AY^2 =BX^2-BY^2$$
$bullet DB'bot AC':$ Since $A'D = DC'$ and $A'B' = C'B'$ we have $$DC'^2-B'C'^2 = DA'^2-B'A'^2 $$
$bullet DB'bot AE:$ Since $AB' = a$, $A'D =asqrt2$, $EB' = asqrt5$ and $DE = asqrt6$ we have $$DE^2-B'E^2 = 6a^2- 5a^2 = 2a^2-a^2 =DA'^2-B'A'^2 $$
and we are done.
answered Jul 22 at 15:47
greedoid
26.1k93473
answered Jul 22 at 15:47
greedoid
26.1k93473
answered Jul 22 at 15:47
greedoid
26.1k93473
answered Jul 22 at 15:47
greedoid
26.1k93473
26.1k93473
add a comment |Â
add a comment |Â
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You seem to be asking two questions: What are other solutions to the problem? and What are generalizations of the problem? These should be posted separately. (You could/should provide links from one to the other to provide context.) BTW: If you want "new" solutions, then you should post the solutions you know so that answerers don't duplicate the effort.
– Blue
Jul 22 at 13:11
goo.gl/images/1iAyQB here is an image of the cube
– Andrei Cataron
Jul 22 at 13:16
About the solutions I have, I'm afraid that if one sees them, it may limit the vision towards others. What do you think?
– Andrei Cataron
Jul 22 at 13:18
1
Math.SE is a question-and-answer site, intended for asking about things you don't know. Therefore, you should post what you know. (You can "hide" solutions by typing
>!in front of the text.) It's no fun for answerers when they devise a clever solution, and go through the trouble of posting it with properly-formatted mathematical expressions and fancy diagrams, only to have the answerer say, "Oh, I already know that one." That's just a waste of everyone's time.– Blue
Jul 22 at 13:36
1
I will edit them in right away.
– Andrei Cataron
Jul 22 at 14:01