Difference between revisions of "2019 AMC 10B Problems/Problem 15"

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Revision as of 04:33, 17 February 2019

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,

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

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions

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