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

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Dividing both sides by <math>\sqrt{3}</math>, we get
 
Dividing both sides by <math>\sqrt{3}</math>, we get
  
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<math>x^2 = \frac{2}{\sqrt{3}}</math>  <math>\rightarrow</math> <math>w^2 = \frac{2\sqrt{3}}{3}</math>
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<math>w^2 = \frac{2}{\sqrt{3}}</math>  <math>\rightarrow</math> <math>w^2 = \frac{2\sqrt{3}}{3}</math>
  
 
Since we want <math>w^2\sqrt{7}</math>, multiplying both sides by <math>\sqrt{7}</math> gets us
 
Since we want <math>w^2\sqrt{7}</math>, multiplying both sides by <math>\sqrt{7}</math> gets us

Revision as of 23:38, 14 February 2019

Problem

Two right triangles, $T_1$ and $T_2$, have areas of 1 and 2, respectively. 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. What is the product of the third side lengths 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,

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

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 $w^2\sqrt{7}$, multiplying both sides by $\sqrt{7}$ gets us

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

Since we want $(wz)^2$, we square the expression giving us

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

-jason168

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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