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

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== Solution ==
 
== Solution ==
20, I'll do the rest later
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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 ==

Revision as of 02:18, 29 January 2019

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

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 (ProblemsAnswer KeyResources)
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