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

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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>?
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<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}}

Latest revision as of 04:13, 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
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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