Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See Also)
(→‎Solution)
Line 6: Line 6:
 
<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>
  
−
==Solution==
+
==Solution 1==
  
 
First of all, name the two sides which are congruent to 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> was 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>.
 
First of all, name the two sides which are congruent to 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> was 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>.
Line 20: Line 20:
 
Since <math>y = \frac{4}{x}</math>, we get <math>y^{4} = \frac{256}{12} = \frac{64}{3}</math>.
 
Since <math>y = \frac{4}{x}</math>, we get <math>y^{4} = \frac{256}{12} = \frac{64}{3}</math>.
  
−
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)}}</math>.
+
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>.
 +
 
 +
==Solution 2==
 +
Like Solution 1, we have <math>\frac{xy}{2} = 2</math> and <math>\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1</math>.
 +
 
 +
Squaring both equations yields <math>x^2y^2=16</math> and <math>x^2(y^2-x^2)=4</math>.
 +
 
 +
Let <math>a = x^2</math> and <math>b = y^2</math>. Then <math>b = \frac{16}{a}</math>, and <math>a(\frac{16}{a}-a)=4 \implies 16 - a^2 = 4 \implies a = 2\sqrt3</math>, so <math>b = \frac{16}{2\sqrt3} = \frac{8\sqrt3}{3}</math>.
 +
 
 +
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>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 17:44, 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 1

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 $\boxed{\textbf{(A) }\frac{28}{3}}$.

Solution 2

Like 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 $a(\frac{16}{a}-a)=4 \implies 16 - a^2 = 4 \implies a = 2\sqrt3$, 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 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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_15&oldid=103169"