2008 UNCO Math Contest II Problems/Problem 3: Difference between revisions
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== Solution == | |||
== Solution 1 == | |||
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>. | |||
The problem says that this is equal to <math>200</math>, so we have <math>a^2+b^2=200</math> | |||
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>. | |||
== 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?" | 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 <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>. | 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>. | ||
Latest revision as of 22:27, 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 
, 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]](http://latex.artofproblemsolving.com/3/3/e/33e59fc38b33e85160f8d9df01198f561a06c879.png)
Solution 1
Let the leg of the longer right triangle have length 
and the shorter one have length 
, so that the area of the four triangles is 
.
The problem says that this is equal to , so we have 
By the Pythagorean theorem, the length of the rectangle is 
and the width is 
, so the length of the rectangle's diagonal is 
. Since , this is simply 
.
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 
, and because the hypotenuse of the 45-45-90 triangle formed by the diagonal is 
*side length, we see that the diagonal of the square and therefore the diagonal of the rectangle is 
or 
.
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 | ||
