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"

(→‎Solution 2)
m (→‎Solution 3)
 
(24 intermediate revisions by 14 users not shown)
Line 2: Line 2:
 
==Problem==
 
==Problem==
  
−
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>?
+
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) sides 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>
  
−
==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, 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>.
  
 
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>.
 
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>.
  
−
We have two equations: <math>\frac{xy}{2} = 2</math> and <math>\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1</math>.
+
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>.
  
 
This means that <math>y = \frac{4}{x}</math> and that <math>\frac{4}{x^{2}} = y^{2} - x^{2}</math>.
 
This means that <math>y = \frac{4}{x}</math> and that <math>\frac{4}{x^{2}} = y^{2} - x^{2}</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==
 
==Solution 2==
 +
Like in Solution 1, we have <math>\frac{xy}{2} = 2</math> and <math>\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1</math>.
  
−
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.  
+
Squaring both equations yields <math>x^2y^2=16</math> and <math>x^2(y^2-x^2)=4</math>.
  
−
So, <math>\overline{AB}</math> ≅ <math>\overline{EF}</math>, call this length <math>x</math>, and <math>\overline{BC}</math> ≅ <math>\overline{DF}</math>, call this length <math>y</math>
+
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>.
  
−
Additionally, call the length <math>\overline{AC}</math> <math>z</math>, and call the length <math>\overline{DE}</math> <math>w</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>.
  
−
Recapping our variables, we have <math>\overline{AB}</math> = <math>\overline{EF}</math> = <math>x</math>, <math>\overline{BC}</math> = <math>\overline{DF}</math> = <math>y</math>, <math>\overline{AC}</math> = <math>z</math>, and <math>\overline{DE}</math> = <math>w</math>
+
==Solution 3==
  
−
We are given that <math>[ABC] = 1</math> and <math>[EDF] = 2</math>
+
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>.
  
−
Since area = <math>\frac{bh}{2}</math>, this gives <math>\frac{xy}{2}</math> = 1 and <math>\frac{xw}{2} = 2</math>
+
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>.
  
−
Dividing the two equations, we get <math>\frac{xy}{xw}</math> = <math>\frac{y}{w} = 2</math>
+
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>.
  
−
From this, we get <math>y = 2w</math>
+
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>.
  
−
We see that △EDF is a <math>30-60-90</math> right triangle, meaning that <math>x = w\sqrt{3}</math>
+
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>.
  
−
In △ABC, <math>x</math> and <math>y</math> are the legs. By the Pythagorean Theorem,
+
[Note: there is a mismatch of variables near the beginning that someone can fix: xw/2 is the area of the small triangle, which is actually the 30-60-90.]
  
−
<math>(w\sqrt{3})^2 + (2w)^2 = z^2</math>  <math>\rightarrow</math>  <math>3w^2 + 4w^2 = z^2</math>  <math>\rightarrow</math>  <math>7w^2 = x^2</math>  <math>\rightarrow</math>  <math>x = w\sqrt{7}</math>
+
==Video Solution==
 +
https://youtu.be/mXvetCMMzpU
  
−
The question asks for the square of the product of the third side lengths of each triangle, which is <math>(wz)^2</math>
+
~IceMatrix
−
 
 
−
Using substitution, we see that <math>wz</math> = <math>(w)(w\sqrt{7}</math>) = <math>w^2\sqrt{7}</math>
 
−
 
 
−
We know <math>\frac{xw}{2} = 1</math>  <math>\rightarrow</math>  <math>\frac{(w)(w\sqrt{3})}{2} =1</math>  <math>\rightarrow</math> <math>(w)(w\sqrt{3}) = 2</math>  <math>\rightarrow</math> <math>(w^2\sqrt{3}) = 2</math>
 
−
 
 
−
Dividing both sides by <math>\sqrt{3}</math>, we get
 
−
 
 
−
<math>x^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
 
−
 
 
−
<math>w^2\sqrt{7} = \frac{2\sqrt{21}}{3}</math>
 
−
 
 
−
Since we want <math>(wz)^2</math>, we square the expression giving us
 
−
 
 
−
<math>(\frac{2\sqrt{21}}{3})^2 = \frac{4*21}{9} = \frac{84}{9} = \frac{28}{3}</math> <math>\rightarrow</math> <math>\boxed{\textbf{(A)}}</math>
 
−
 
 
−
-jason168
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2019|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2019|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:59, 22 March 2023

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

[Note: there is a mismatch of variables near the beginning that someone can fix: xw/2 is the area of the small triangle, which is actually the 30-60-90.]

Video Solution

https://youtu.be/mXvetCMMzpU

~IceMatrix

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=191201"