2012 AMC 8 Problems/Problem 25

Problem

A square with area $4$ is inscribed in a square with area $5$ , with each vertex of the smaller square on a side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length $a$ , and the other of length $b$ . What is the value of $ab$ ?

$[asy] draw((0,2)--(2,2)--(2,0)--(0,0)--cycle); draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle); label("$a$",(-0.1,0.15)); label("$b$",(-0.1,1.15));[/asy]$

$\textbf{(A)}\hspace{.05in}\frac{1}5\qquad\textbf{(B)}\hspace{.05in}\frac{2}5\qquad\textbf{(C)}\hspace{.05in}\frac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4$

Solution 1

The total area of the four congruent triangles formed by the squares is $5-4 = 1$ . Therefore, the area of one of these triangles is $\frac{1}{4}$ . The height of one of these triangles is $a$ and the base is $b$ . Using the formula for area of the triangle, we have $\frac{ab}{2} = \frac{1}{4}$ . Multiply by $2$ on both sides to find that the value of $ab$ is $\boxed{\textbf{(C)}\ \frac{1}2}$ .

Solution 2 (Algebra)

We see that we want $ab$ , so instead of solving for $a,b$ , we find a way to get an expression with $ab$ .

By Triple Perpendicularity Model,

all four triangles are congruent.

By Pythagorean's Theorem,

$\sqrt{a^2+b^2} = \sqrt{4}$

Thus, $\sqrt{a^2+b^2} = 2$

As $a+b=\sqrt{5}$ ,

$a^2+2ab+b^2 = 5$

So, $\sqrt{5-2ab} = 2$

Simplifying,

$5-2ab = 4$

$-2ab=-1$

$ab=\frac{1}{2}$ or $\boxed{\textbf{(C)} \frac{1}{2}}$

Video Solution 2

https://youtu.be/MhxGq1sSA6U ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=2

~ pi_is_3.14

2012 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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