2019 AMC 10B Problems/Problem 15

Problem

Right triangles $T_1$ and $T_2$ , have areas of 1 and 2, respectively. A side of $T_1$ is congruent to a side of $T_2$ , and a different side of $T_1$ is congruent to a different side of $T_2$ . What is the square of the product of the lengths of the other (third) side of $T_1$ and $T_2$ ?

$\textbf{(A) }\frac{28}{3}\qquad\textbf{(B) }10\qquad\textbf{(C) }\frac{32}{3}\qquad\textbf{(D) }\frac{34}{3}\qquad\textbf{(E) }12$

Solution

First of all, name the two sides which are congruent to be $x$ and $y$ , where $y > x$ . The only way that the conditions of the problem can be satisfied is if $x$ was the shorter leg of $T_{2}$ and the longer leg of $T_{1}$ , and $y$ is the longer leg of $T_{2}$ and the hypotenuse of $T_{1}$ .

Notice that this means the value we are looking for is the square of $\sqrt{x^{2}+y^{2}} \cdot \sqrt{y^{2}-x^{2}} = \sqrt{y^{4}-x^{4}}$ , which is just $y^{4}-x^{4}$ .

We have two equations: $\frac{xy}{2} = 2$ and $\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1$ .

This means that $y = \frac{4}{x}$ and that $\frac{4}{x^{2}} = y^{2} - x^{2}$ .

Taking the second equation, we get $x^{2}y^{2} - x^{4} = 4$ , so since $xy = 4$ , $x^{4} = 12$ .

Since $y = \frac{4}{x}$ , we get $y^{4} = \frac{256}{12} = \frac{64}{3}$ .

The value we are looking for is just $y^{4}-x^{4} = \frac{64-36}{3} = \frac{28}{3}$ so the answer is $\boxed{\textbf{(A)}}$ .

Solution 2

First, construct right triangles △ABC and △EDF, with △ABC being the smaller triangle. We are given that one side length of one triangle is congruent to a different side length in the other, and another side length of the first triangle is congruent to yet another side length in the other.

So, $\overline{AB}$ ≅ $\overline{EF}$ , call this length $x$ , and $\overline{BC}$ ≅ $\overline{DF}$ , call this length $y$

Additionally, call the length $\overline{AC}$ $z$ , and call the length $\overline{DE}$ $w$

Recapping our variables, we have $\overline{AB}$ = $\overline{EF}$ = $x$ , $\overline{BC}$ = $\overline{DF}$ = $y$ , $\overline{AC}$ = $z$ , and $\overline{DE}$ = $w$

We are given that $[ABC] = 1$ and $[EDF] = 2$

Since area = $\frac{bh}{2}$ , this gives $\frac{xy}{2} = 1$ and $\frac{xw}{2} = 2$

Dividing the two equations, we get $\frac{xy}{xw}$ = $\frac{y}{w} = 2$

From this, we get $y = 2w$

We see that △EDF is a $30-60-90$ right triangle, meaning that $x = w\sqrt{3}$

In △ABC, $x$ and $y$ are the legs. By the Pythagorean Theorem,

$(w\sqrt{3})^2 + (2w)^2 = z^2$ $\rightarrow$ $3w^2 + 4w^2 = z^2$ $\rightarrow$ $7w^2 = z^2$ $\rightarrow$ $w\sqrt{7} = z$

The question asks for the square of the product of the third side lengths of each triangle, which is $(wz)^2$

Using substitution, we see that $wz$ = $(w)(w\sqrt{7}$ ) = $w^2\sqrt{7}$

We know $\frac{xw}{2} = 1$ $\rightarrow$ $\frac{(w)(w\sqrt{3})}{2} =1$ $\rightarrow$ $(w)(w\sqrt{3}) = 2$ $\rightarrow$ $(w^2\sqrt{3}) = 2$

Dividing both sides by $\sqrt{3}$ , we get

$w^2 = \frac{2}{\sqrt{3}}$ $\rightarrow$ $w^2 = \frac{2\sqrt{3}}{3}$

Since we want $(w^2\sqrt{7})^2$ , multiplying both sides by $\sqrt{7}$ gets us

$w^2\sqrt{7} = \frac{2\sqrt{21}}{3}$

Squaring this,

$(\frac{2\sqrt{21}}{3})^2 = \frac{4*21}{9} = \frac{84}{9} = \frac{28}{3}$ $\rightarrow$ $\boxed{\textbf{(A)}}$ .

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2019 AMC 10B Problems/Problem 15

Contents

Problem

Solution

Solution 2

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