Segment measure in right triangles

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The $triangle ABO$ is rectangle in $O$, $OA=a$ and $OB=b$, if $AP=PQ=QB=x$, then $x=?$



enter image description here



I tried pythagoras in $triangle POQ$, because $OP=a-x$, $OQ=b-x$ and $PQ=x$ and got a messy equation that i can't solve. That's what i tried for now.




geometry euclidean-geometry triangle




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  • Edited. I put what i tried.
    – Rodrigo Pizarro
    Jul 15 at 2:55










  • Show us the "messy" equation. Otherwise we don't really know what you tried and we don't know how close you were to a solution.
    – David K
    Jul 15 at 3:01















up vote
2
down vote

favorite












The $triangle ABO$ is rectangle in $O$, $OA=a$ and $OB=b$, if $AP=PQ=QB=x$, then $x=?$



enter image description here



I tried pythagoras in $triangle POQ$, because $OP=a-x$, $OQ=b-x$ and $PQ=x$ and got a messy equation that i can't solve. That's what i tried for now.




geometry euclidean-geometry triangle




share|cite|improve this question





















  • Edited. I put what i tried.
    – Rodrigo Pizarro
    Jul 15 at 2:55










  • Show us the "messy" equation. Otherwise we don't really know what you tried and we don't know how close you were to a solution.
    – David K
    Jul 15 at 3:01













up vote
2
down vote

favorite









up vote
2
down vote

favorite











The $triangle ABO$ is rectangle in $O$, $OA=a$ and $OB=b$, if $AP=PQ=QB=x$, then $x=?$



enter image description here



I tried pythagoras in $triangle POQ$, because $OP=a-x$, $OQ=b-x$ and $PQ=x$ and got a messy equation that i can't solve. That's what i tried for now.




geometry euclidean-geometry triangle




share|cite|improve this question













The $triangle ABO$ is rectangle in $O$, $OA=a$ and $OB=b$, if $AP=PQ=QB=x$, then $x=?$



enter image description here



I tried pythagoras in $triangle POQ$, because $OP=a-x$, $OQ=b-x$ and $PQ=x$ and got a messy equation that i can't solve. That's what i tried for now.





geometry euclidean-geometry triangle





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 15 at 2:54
























asked Jul 15 at 2:44









Rodrigo Pizarro

696117




696117











  • Edited. I put what i tried.
    – Rodrigo Pizarro
    Jul 15 at 2:55










  • Show us the "messy" equation. Otherwise we don't really know what you tried and we don't know how close you were to a solution.
    – David K
    Jul 15 at 3:01

















  • Edited. I put what i tried.
    – Rodrigo Pizarro
    Jul 15 at 2:55










  • Show us the "messy" equation. Otherwise we don't really know what you tried and we don't know how close you were to a solution.
    – David K
    Jul 15 at 3:01
















Edited. I put what i tried.
– Rodrigo Pizarro
Jul 15 at 2:55




Edited. I put what i tried.
– Rodrigo Pizarro
Jul 15 at 2:55












Show us the "messy" equation. Otherwise we don't really know what you tried and we don't know how close you were to a solution.
– David K
Jul 15 at 3:01





Show us the "messy" equation. Otherwise we don't really know what you tried and we don't know how close you were to a solution.
– David K
Jul 15 at 3:01











1 Answer
1






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oldest

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2
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Of course you can solve it easily, this being a quadratic. Some terms even cancel out nicely:
$$(a-x)^2+(b-x)^2=x^2$$
$$x^2-2(a+b)x+a^2+b^2=0$$
$$x=frac2(a+b)pmsqrt4(a+b)^2-4(a^2+b^2)2$$
$$x=a+b-sqrt2ab$$
We reject $+$ in the last line because $PQ$ can never be longer than $a+b$.






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  • That was the trick, thanks.
    – Rodrigo Pizarro
    Jul 15 at 3:11










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Of course you can solve it easily, this being a quadratic. Some terms even cancel out nicely:
$$(a-x)^2+(b-x)^2=x^2$$
$$x^2-2(a+b)x+a^2+b^2=0$$
$$x=frac2(a+b)pmsqrt4(a+b)^2-4(a^2+b^2)2$$
$$x=a+b-sqrt2ab$$
We reject $+$ in the last line because $PQ$ can never be longer than $a+b$.






share|cite|improve this answer





















  • That was the trick, thanks.
    – Rodrigo Pizarro
    Jul 15 at 3:11














up vote
2
down vote



accepted










Of course you can solve it easily, this being a quadratic. Some terms even cancel out nicely:
$$(a-x)^2+(b-x)^2=x^2$$
$$x^2-2(a+b)x+a^2+b^2=0$$
$$x=frac2(a+b)pmsqrt4(a+b)^2-4(a^2+b^2)2$$
$$x=a+b-sqrt2ab$$
We reject $+$ in the last line because $PQ$ can never be longer than $a+b$.






share|cite|improve this answer





















  • That was the trick, thanks.
    – Rodrigo Pizarro
    Jul 15 at 3:11












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Of course you can solve it easily, this being a quadratic. Some terms even cancel out nicely:
$$(a-x)^2+(b-x)^2=x^2$$
$$x^2-2(a+b)x+a^2+b^2=0$$
$$x=frac2(a+b)pmsqrt4(a+b)^2-4(a^2+b^2)2$$
$$x=a+b-sqrt2ab$$
We reject $+$ in the last line because $PQ$ can never be longer than $a+b$.






share|cite|improve this answer













Of course you can solve it easily, this being a quadratic. Some terms even cancel out nicely:
$$(a-x)^2+(b-x)^2=x^2$$
$$x^2-2(a+b)x+a^2+b^2=0$$
$$x=frac2(a+b)pmsqrt4(a+b)^2-4(a^2+b^2)2$$
$$x=a+b-sqrt2ab$$
We reject $+$ in the last line because $PQ$ can never be longer than $a+b$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 15 at 3:02









Parcly Taxel

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  • That was the trick, thanks.
    – Rodrigo Pizarro
    Jul 15 at 3:11
















  • That was the trick, thanks.
    – Rodrigo Pizarro
    Jul 15 at 3:11















That was the trick, thanks.
– Rodrigo Pizarro
Jul 15 at 3:11




That was the trick, thanks.
– Rodrigo Pizarro
Jul 15 at 3:11












 

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