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Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 3"

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== Solution ==
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20, I'll do the rest later
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== Solution 1 ==
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Let the leg of the longer right triangle have length <math>a</math> and the shorter one have length <math>b</math>, so that the area of the four triangles is <math>\frac{1}{2}a\cdot a+\frac{1}{2}b\cdot b+\frac{1}{2}a\cdot a+\frac{1}{2}b\cdot b = a^2+b^2</math>.
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The problem says that this is equal to <math>200</math>, so we have <math>a^2+b^2=200</math>
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By the [[Pythagorean theorem]], the length of the rectangle is <math>\sqrt{a^2+a^2}=a\sqrt{2}</math> and the width is <math>\sqrt{b^2+b^2}=b\sqrt{2}</math>, so the length of the rectangle's diagonal is <math>\sqrt{(\sqrt{2}a)^2+(\sqrt{2}b)^2} = \sqrt{2(a^2+b^2)}</math>. Since <math>a^2+b^2=200</math>, this is simply <math>\sqrt{2\cdot 200} = \boxed{20}</math>.
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== Solution 2==
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Without loss of generality, squeeze the rectangle into a line that becomes the diagonal of the square. 2 of the triangles approach 0 area as the rectangle approaches a line and the diagonal of the rectangle approaches the line, so we can treat this as a question of "What is the length of the diagonal of a square of area 200?"
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We see that the side of the square must be <math>\sqrt{200}</math>, and because the hypotenuse of the 45-45-90 triangle formed by the diagonal is <math>\sqrt{2}</math>*side length, we see that the diagonal of the square and therefore the diagonal of the rectangle is <math>\sqrt{400}</math> or <math>\boxed{20}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 23:27, 13 January 2024

Problem

A rectangle is inscribed in a square creating four isosceles right triangles. If the total area of these four triangles is $200$, what is the length of the diagonal of the rectangle?

[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); draw((0,2/3)--(2/3,0)--(1,1/3)--(1/3,1)--cycle,black); [/asy]


Solution 1

Let the leg of the longer right triangle have length $a$ and the shorter one have length $b$, so that the area of the four triangles is $\frac{1}{2}a\cdot a+\frac{1}{2}b\cdot b+\frac{1}{2}a\cdot a+\frac{1}{2}b\cdot b = a^2+b^2$. The problem says that this is equal to $200$, so we have $a^2+b^2=200$

By the Pythagorean theorem, the length of the rectangle is $\sqrt{a^2+a^2}=a\sqrt{2}$ and the width is $\sqrt{b^2+b^2}=b\sqrt{2}$, so the length of the rectangle's diagonal is $\sqrt{(\sqrt{2}a)^2+(\sqrt{2}b)^2} = \sqrt{2(a^2+b^2)}$. Since $a^2+b^2=200$, this is simply $\sqrt{2\cdot 200} = \boxed{20}$.

Solution 2

Without loss of generality, squeeze the rectangle into a line that becomes the diagonal of the square. 2 of the triangles approach 0 area as the rectangle approaches a line and the diagonal of the rectangle approaches the line, so we can treat this as a question of "What is the length of the diagonal of a square of area 200?" We see that the side of the square must be $\sqrt{200}$, and because the hypotenuse of the 45-45-90 triangle formed by the diagonal is $\sqrt{2}$*side length, we see that the diagonal of the square and therefore the diagonal of the rectangle is $\sqrt{400}$ or $\boxed{20}$.

See Also

2008 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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