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

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Two right triangles, <math>T_1</math> and <math>T_2</math>, 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 <math>T_1</math> and <math>T_2</math>?
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==Problem==
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Right triangles <math>T_1</math> and <math>T_2</math>, have areas of 1 and 2, respectively. A side of <math>T_1</math> is congruent to a side of <math>T_2</math>, and a different side of <math>T_1</math> is congruent to a different side of <math>T_2</math>. What is the square of the product of the lengths of the other (third) side of <math>T_1</math> and <math>T_2</math>?
  
 
<math>\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</math>
 
<math>\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</math>
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==Solution 1==
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First of all, let the two sides which are congruent be <math>x</math> and <math>y</math>, where <math>y > x</math>. The only way that the conditions of the problem can be satisfied is if <math>x</math> is the shorter leg of <math>T_{2}</math> and the longer leg of <math>T_{1}</math>, and <math>y</math> is the longer leg of <math>T_{2}</math> and the hypotenuse of <math>T_{1}</math>.
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Notice that this means the value we are looking for is the square of <math>\sqrt{x^{2}+y^{2}} \cdot \sqrt{y^{2}-x^{2}} = \sqrt{y^{4}-x^{4}}</math>, which is just <math>y^{4}-x^{4}</math>.
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The area conditions give us two equations: <math>\frac{xy}{2} = 2</math> and <math>\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1</math>.
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This means that <math>y = \frac{4}{x}</math> and that <math>\frac{4}{x^{2}} = y^{2} - x^{2}</math>.
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Taking the second equation, we get <math>x^{2}y^{2} - x^{4} = 4</math>, so since <math>xy = 4</math>, <math>x^{4} = 12</math>.
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Since <math>y = \frac{4}{x}</math>, we get <math>y^{4} = \frac{256}{12} = \frac{64}{3}</math>.
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The value we are looking for is just <math>y^{4}-x^{4} = \frac{64-36}{3} = \frac{28}{3}</math> so the answer is <math>\boxed{\textbf{(A) }\frac{28}{3}}</math>.
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==Solution 2==
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Like in Solution 1, we have <math>\frac{xy}{2} = 2</math> and <math>\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1</math>.
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Squaring both equations yields <math>x^2y^2=16</math> and <math>x^2(y^2-x^2)=4</math>.
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Let <math>a = x^2</math> and <math>b = y^2</math>. Then <math>b = \frac{16}{a}</math>, and <math>a\left(\frac{16}{a}-a\right)=4 \implies 16 - a^2 = 4 \implies a = 2\sqrt3</math>, so <math>b = \frac{16}{2\sqrt3} = \frac{8\sqrt3}{3}</math>.
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We are looking for the value of <math>y^4 - x^4 = b^2 - a^2</math>, so the answer is <math>\frac{64}{3} - 12 = \boxed{\textbf{(A) }\frac{28}{3}}</math>.
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==Solution 3==
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Firstly, let the right triangles be <math>\triangle ABC</math> and <math>\triangle EDF</math>, with <math>\triangle ABC</math> being the smaller triangle. As in Solution 1, let <math>\overline{AB} = \overline{EF} = x</math> and <math>\overline{BC} = \overline{DF} = y</math>. Additionally, let <math>\overline{AC} = z</math> and <math>\overline{DE} = w</math>.
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We are given that <math>[ABC] = 1</math> and <math>[EDF] = 2</math>, so using <math>\text{area} = \frac{bh}{2}</math>, we have <math>\frac{xy}{2} = 1</math> and <math>\frac{xw}{2} = 2</math>. Dividing the two equations, we get <math>\frac{xy}{xw}</math> = <math>\frac{y}{w} = 2</math>, so <math>y = 2w</math>.
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Thus <math>\triangle EDF</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> right triangle, meaning that <math>x = w\sqrt{3}</math>. Now by the Pythagorean Theorem in <math>\triangle ABC</math>, <math>\left(w\sqrt{3}\right)^2 + \left(2w\right)^2 = z^2 \Rightarrow 3w^2 + 4w^2 = z^2 \Rightarrow 7w^2 = z^2 \Rightarrow w\sqrt{7} = z</math>.
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The problem requires the square of the product of the third side lengths of each triangle, which is <math>(wz)^2</math>. By substitution, we see that <math>wz</math> = <math>\left(w\right)\left(w\sqrt{7}\right) = w^2\sqrt{7}</math>. We also know <math>\frac{xw}{2} = 1 \Rightarrow\frac{(w)\left(w\sqrt{3}\right)}{2} =1 \Rightarrow (w)\left(w\sqrt{3}\right) = 2 \Rightarrow w^2\sqrt{3} = 2 \Rightarrow w^2 = \frac{2\sqrt{3}}{3}</math>.
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Since we want <math>\left(w^2\sqrt{7}\right)^2</math>, multiplying both sides by <math>\sqrt{7}</math> gets us <math>w^2\sqrt{7} = \frac{2\sqrt{21}}{3}</math>. Now squaring gives <math>\left(\frac{2\sqrt{21}}{3}\right)^2 = \frac{4*21}{9} = \boxed{\textbf{(A) }\frac{28}{3}}</math>.
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==Video Solution==
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https://youtu.be/mXvetCMMzpU
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~IceMatrix
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==See Also==
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{{AMC10 box|year=2019|ab=B|num-b=14|num-a=16}}
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{{MAA Notice}}

Revision as of 05:13, 28 January 2020

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 1

First of all, let the two sides which are congruent be $x$ and $y$, where $y > x$. The only way that the conditions of the problem can be satisfied is if $x$ is 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 $y^{4}-x^{4}$.

The area conditions give us 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 $\boxed{\textbf{(A) }\frac{28}{3}}$.

Solution 2

Like in Solution 1, we have $\frac{xy}{2} = 2$ and $\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1$.

Squaring both equations yields $x^2y^2=16$ and $x^2(y^2-x^2)=4$.

Let $a = x^2$ and $b = y^2$. Then $b = \frac{16}{a}$, and $a\left(\frac{16}{a}-a\right)=4 \implies 16 - a^2 = 4 \implies a = 2\sqrt3$, so $b = \frac{16}{2\sqrt3} = \frac{8\sqrt3}{3}$.

We are looking for the value of $y^4 - x^4 = b^2 - a^2$, so the answer is $\frac{64}{3} - 12 = \boxed{\textbf{(A) }\frac{28}{3}}$.

Solution 3

Firstly, let the right triangles be $\triangle ABC$ and $\triangle EDF$, with $\triangle ABC$ being the smaller triangle. As in Solution 1, let $\overline{AB} = \overline{EF} = x$ and $\overline{BC} = \overline{DF} = y$. Additionally, let $\overline{AC} = z$ and $\overline{DE} = w$.

We are given that $[ABC] = 1$ and $[EDF] = 2$, so using $\text{area} = \frac{bh}{2}$, we have $\frac{xy}{2} = 1$ and $\frac{xw}{2} = 2$. Dividing the two equations, we get $\frac{xy}{xw}$ = $\frac{y}{w} = 2$, so $y = 2w$.

Thus $\triangle EDF$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ right triangle, meaning that $x = w\sqrt{3}$. Now by the Pythagorean Theorem in $\triangle ABC$, $\left(w\sqrt{3}\right)^2 + \left(2w\right)^2 = z^2 \Rightarrow 3w^2 + 4w^2 = z^2 \Rightarrow 7w^2 = z^2 \Rightarrow w\sqrt{7} = z$.

The problem requires the square of the product of the third side lengths of each triangle, which is $(wz)^2$. By substitution, we see that $wz$ = $\left(w\right)\left(w\sqrt{7}\right) = w^2\sqrt{7}$. We also know $\frac{xw}{2} = 1 \Rightarrow\frac{(w)\left(w\sqrt{3}\right)}{2} =1 \Rightarrow (w)\left(w\sqrt{3}\right) = 2 \Rightarrow w^2\sqrt{3} = 2 \Rightarrow w^2 = \frac{2\sqrt{3}}{3}$.

Since we want $\left(w^2\sqrt{7}\right)^2$, multiplying both sides by $\sqrt{7}$ gets us $w^2\sqrt{7} = \frac{2\sqrt{21}}{3}$. Now squaring gives $\left(\frac{2\sqrt{21}}{3}\right)^2 = \frac{4*21}{9} = \boxed{\textbf{(A) }\frac{28}{3}}$.

Video Solution

https://youtu.be/mXvetCMMzpU

~IceMatrix

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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