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 "2008 AMC 10A Problems/Problem 18"
m (→‎Solution 8)
 
(15 intermediate revisions by 3 users not shown)
Line 26: Line 26:
  
 
=== Solution 2 ===
 
=== Solution 2 ===
−
From the formula <math>A = rs</math>, where <math>A</math> is the area of a triangle, <math>r</math> is its [[inradius]], and <math>s</math> is the [[semiperimeter]], we can find that <math>r = \frac{20}{32/2} = \frac{5}{4}</math>. It is known that in a right triangle, <math>r = s - h</math>, where <math>h</math> is the hypotenuse, so <math>h = 16 - \frac{5}{4} = \frac{59}{4}</math> (B).
+
From the formula <math>A = rs</math>, where <math>A</math> is the area of a triangle, <math>r</math> is its [[inradius]], and <math>s</math> is the [[semiperimeter]], we can find that <math>r = \frac{20}{32/2} = \frac{5}{4}</math>. In a right triangle, <math>r = s - h</math>, where <math>h</math> is the hypotenuse, so <math>h = 16 - \frac{5}{4} = \frac{59}{4}</math> <math>\mathrm{(B)}</math>.
 +
 
 +
'''Note:''' If the reader is unfamiliar with the inradius being equal to the semiperimeter minus the hypotenuse for a right triangle, it should not be too difficult to prove the relationship for themselves.~mobius247
 +
 
 +
'''Why <math>r=s-h</math>.''' Draw a right triangle and inscribe a circle. Connect the center of the circle with tangency points; there are 3, one from each side of the triangle. Now, connect the center with each vertice and you will see 3 pairs of congruent triangles by HL (Hypotenuse-Leg Congruence). From these congruent triangles, mark equal lengths and you will find that <math>r=s-h</math>.  
 +
 
 +
~BakedPotato66
  
 
=== Solution 3 ===
 
=== Solution 3 ===
Line 47: Line 53:
 
\end{align*}</cmath></center>
 
\end{align*}</cmath></center>
  
−
Further simplification yields the result of <math>\frac{59}{4}</math>.
+
Further simplification yields the result of <math>\frac{59}{4} \rightarrow \mathrm{(B)}</math>.
  
 
=== Solution 4 ===
 
=== Solution 4 ===
−
Let <math>a</math> and <math>b</math> be the legs of the triangle, and <math>c</math> the hypotenuse.
+
Let <math>a</math> and <math>b</math> be the legs of the triangle and <math>c</math> the hypotenuse.
  
 
Since the area is 20, we have <math>\frac{1}{2}ab = 20 => ab=40</math>.
 
Since the area is 20, we have <math>\frac{1}{2}ab = 20 => ab=40</math>.
Line 71: Line 77:
 
64c = 944 \\\\
 
64c = 944 \\\\
 
c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4} </math>. </center>
 
c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4} </math>. </center>
−
The answer is choice (B).
+
The answer is choice <math>(B)</math>.
  
 
=== Solution 5 ===
 
=== Solution 5 ===
−
Let <math>a</math>, <math>b</math>, and <math>c</math> be the sides of the triangle, with <math>c</math> as the hypotenuse.
+
Let <math>a</math>, <math>b</math>, and <math>c</math> be the sides of the triangle with <math>c</math> as the hypotenuse.
  
 
We know that <math>a + b + c =32</math>.
 
We know that <math>a + b + c =32</math>.
Line 80: Line 86:
 
According to the Pythagorean Theorem, we have <math>a^2 + b^2 = c^2</math>.
 
According to the Pythagorean Theorem, we have <math>a^2 + b^2 = c^2</math>.
  
−
We also know that <math>ab</math> = 40, since the area of the triangle is 20.
+
We also know that <math>ab = 40</math>, since the area of the triangle is <math>20</math>.
  
 
We substitute <math>2ab</math> into <math>a^2 + b^2 = c^2</math> to get <math>(a+b)^2 = c^2 + 80</math>.
 
We substitute <math>2ab</math> into <math>a^2 + b^2 = c^2</math> to get <math>(a+b)^2 = c^2 + 80</math>.
Line 86: Line 92:
 
Moving the <math>c^2</math> to the left, we again rewrite to get <math>(a+b+c)(a+b-c) = 80</math>.
 
Moving the <math>c^2</math> to the left, we again rewrite to get <math>(a+b+c)(a+b-c) = 80</math>.
  
−
We substitute our value of 32 for <math>a+b+c</math> twice into our equation and subtract to get <math>a + b = \frac{69}{4}</math>.
+
We substitute our value of <math>32</math> for <math>a+b+c</math> twice into our equation and subtract to get <math>a + b = \frac{69}{4}</math>.
−
 
 
−
Finally, subtracting this from our original value of 32, we get <math>\frac{59}{4}</math>, or <math>B</math>.
 
  
 +
Finally, subtracting this from our original value of <math>32</math>, we get <math>\frac{59}{4}</math>, or <math>B</math>.
  
 
=== Solution 6 ===
 
=== Solution 6 ===
Line 109: Line 114:
 
Multiplying both sides by <math>4a</math> gives: <cmath>4a^2 - 69a + 160.</cmath>
 
Multiplying both sides by <math>4a</math> gives: <cmath>4a^2 - 69a + 160.</cmath>
  
−
It's not hard to see that the roots of this equation are <math>a</math> and <math>b</math>. We want the hypotenuse which is <math>\sqrt{a^2 + b^2} = \sqrt{ (a+b)^2 - 2ab}.</math>
+
It can be observed that the roots of this equation are <math>a</math> and <math>b</math>. We want the hypotenuse which is <math>\sqrt{a^2 + b^2} = \sqrt{ (a+b)^2 - 2ab}.</math>
  
 
We can now apply Vieta's Formula which gives: <cmath>c = \sqrt{ \left( \frac{69}{4} \right)^2 - 80} = \boxed{ \frac{59}{4}}.</cmath>
 
We can now apply Vieta's Formula which gives: <cmath>c = \sqrt{ \left( \frac{69}{4} \right)^2 - 80} = \boxed{ \frac{59}{4}}.</cmath>
Line 191: Line 196:
  
 
~mathboy282
 
~mathboy282
 +
 +
=== Solution 8 ===
 +
This solution is very similar to Solution 1, except instead of subtracting <math>a+b</math> from both sides in the first part, we subtract <math>\sqrt{a^2+b^2}</math> from both sides, which gets us:
 +
 +
<cmath>
 +
\begin{align*}
 +
a+b&=32-\sqrt{a^2+b^2}\\
 +
(a+b)^2&=(32-\sqrt{a^2+b^2})^2\\
 +
a^2+2ab+b^2&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}+a^2+b^2\\
 +
2ab&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}.\\
 +
\end{align*}
 +
</cmath>
 +
 +
We know that
 +
<cmath>\frac{1}{2}ab=20</cmath>
 +
so
 +
<cmath>ab=40.</cmath>
 +
 +
We can then substitute <math>40</math> for <math>ab</math> to get us:
 +
 +
<cmath>
 +
\begin{align*}
 +
80&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}\\
 +
2\cdot32\cdot\sqrt{a^2+b^2}&=32^2-80\\
 +
64\cdot\sqrt{a^2+b^2}&=32^2-80\\
 +
\sqrt{a^2+b^2}&=\frac{32^2-80}{64}\\
 +
\end{align*}
 +
</cmath>
 +
 +
We know that <math>\sqrt{a^2+b^2}</math> is the hypotenuse, so we only have to solve the right-hand side now.
 +
 +
<cmath>
 +
\begin{align*}
 +
\sqrt{a^2+b^2}&=\frac{32^2-80}{64}\\
 +
&=\frac{32^2}{64}-\frac{80}{64}\\
 +
&=\frac{32\cdot32}{32\cdot2}-\frac{32\cdot\frac{5}{2}}{32\cdot2}\\
 +
&=\frac{32}{2}-\frac{\frac{5}{2}}{2}\\
 +
&=16-\frac{5}{4}\\
 +
&=\frac{64}{4}-\frac{5}{4}\\
 +
&=\boxed{\mathrm{(B)} \frac{59}{4}}\\
 +
\end{align*}
 +
</cmath>
 +
 +
~zlrara01
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 13:15, 1 August 2022

Problem

A right triangle has perimeter $32$ and area $20$. What is the length of its hypotenuse?

$\mathrm{(A)}\ \frac{57}{4}\qquad\mathrm{(B)}\ \frac{59}{4}\qquad\mathrm{(C)}\ \frac{61}{4}\qquad\mathrm{(D)}\ \frac{63}{4}\qquad\mathrm{(E)}\ \frac{65}{4}$

Solution

Solution 1

Let the legs of the triangle have lengths $a,b$. Then, by the Pythagorean Theorem, the length of the hypotenuse is $\sqrt{a^2+b^2}$, and the area of the triangle is $\frac 12 ab$. So we have the two equations

$a+b+\sqrt{a^2+b^2} = 32 \\\\ \frac{1}{2}ab = 20$

Re-arranging the first equation and squaring,

$\sqrt{a^2+b^2} = 32-(a+b)\\\\ a^2 + b^2 = 32^2 - 64(a+b) + (a+b)^2\\\\ a^2 + b^2 + 64(a+b) = a^2 + b^2 + 2ab + 32^2\\\\ a+b = \frac{2ab+32^2}{64}$

From $(2)$ we have $2ab = 80$, so

$a+b = \frac{80 + 32^2}{64} = \frac{69}{4}.$

The length of the hypotenuse is $p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\boxed{\ \mathrm{(B)}}$.

Solution 2

From the formula $A = rs$, where $A$ is the area of a triangle, $r$ is its inradius, and $s$ is the semiperimeter, we can find that $r = \frac{20}{32/2} = \frac{5}{4}$. In a right triangle, $r = s - h$, where $h$ is the hypotenuse, so $h = 16 - \frac{5}{4} = \frac{59}{4}$ $\mathrm{(B)}$.

Note: If the reader is unfamiliar with the inradius being equal to the semiperimeter minus the hypotenuse for a right triangle, it should not be too difficult to prove the relationship for themselves.~mobius247

Why $r=s-h$. Draw a right triangle and inscribe a circle. Connect the center of the circle with tangency points; there are 3, one from each side of the triangle. Now, connect the center with each vertice and you will see 3 pairs of congruent triangles by HL (Hypotenuse-Leg Congruence). From these congruent triangles, mark equal lengths and you will find that $r=s-h$.

~BakedPotato66

Solution 3

From the problem, we know that

\begin{align*} a+b+c &= 32 \\ 2ab &= 80. \\ \end{align*}

Subtracting $c$ from both sides of the first equation and squaring both sides, we get

\begin{align*} (a+b)^2 &= (32 - c)^2\\ a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\ \end{align*}

Now we substitute in $a^2 + b^2 = c^2$ as well as $2ab = 80$ into the equation to get

\begin{align*} 80 &= 1024 - 64c\\ c &= \frac{944}{64}. \end{align*}

Further simplification yields the result of $\frac{59}{4} \rightarrow \mathrm{(B)}$.

Solution 4

Let $a$ and $b$ be the legs of the triangle and $c$ the hypotenuse.

Since the area is 20, we have $\frac{1}{2}ab = 20 => ab=40$.

Since the perimeter is 32, we have $a + b + c = 32$.

The Pythagorean Theorem gives $c^2 = a^2 + b^2$.

This gives us three equations with three variables:

$ab = 40 \\ a + b + c = 32 \\ c^2 = a^2 + b^2$

Rewrite equation 3 as $c^2 = (a+b)^2 - 2ab$. Substitute in equations 1 and 2 to get $c^2 = (32-c)^2 - 80$.

$c^2 = (32-c)^2 - 80 \\\\ c^2 = 1024 - 64c + c^2 - 80 \\\\ 64c = 944 \\\\ c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4}$.

The answer is choice $(B)$.

Solution 5

Let $a$, $b$, and $c$ be the sides of the triangle with $c$ as the hypotenuse.

We know that $a + b + c =32$.

According to the Pythagorean Theorem, we have $a^2 + b^2 = c^2$.

We also know that $ab = 40$, since the area of the triangle is $20$.

We substitute $2ab$ into $a^2 + b^2 = c^2$ to get $(a+b)^2 = c^2 + 80$.

Moving the $c^2$ to the left, we again rewrite to get $(a+b+c)(a+b-c) = 80$.

We substitute our value of $32$ for $a+b+c$ twice into our equation and subtract to get $a + b = \frac{69}{4}$.

Finally, subtracting this from our original value of $32$, we get $\frac{59}{4}$, or $B$.

Solution 6

Let the legs be $a, b$. Then the hypotenuse is $\sqrt{a^2 + b^2}.$

We know that $ab = 40$ and $a + b + \sqrt{a^2 + b^2} = 32$.

The first equation gives \[b = \frac{40}{a}\] and we can plug this into the second equation, yielding: \[a + \frac{40}{a} + \sqrt{ a^2 + \frac{1600}{a^2}} = 32.\]

Letting $X = a + \frac{40}{a}$, the equation becomes: \[X + \sqrt{X^2 - 80} = 32.\]

We can bring the $X$ to the right side and square which yields: \[X^2 - 80 = (X - 32)^2 = X^2 - 64X + 1024.\]

So, \[-80 = -64X + 1024 \rightarrow X = \frac{69}{4}.\]

Now, we know that $a + \frac{40}{a} = \frac{69}{4}.$

Multiplying both sides by $4a$ gives: \[4a^2 - 69a + 160.\]

It can be observed that the roots of this equation are $a$ and $b$. We want the hypotenuse which is $\sqrt{a^2 + b^2} = \sqrt{ (a+b)^2 - 2ab}.$

We can now apply Vieta's Formula which gives: \[c = \sqrt{ \left( \frac{69}{4} \right)^2 - 80} = \boxed{ \frac{59}{4}}.\]

~conantwiz2023

Solution 7

Let the sides be $a, b, c$ where $a$ and $b$ are the legs and $c$ is the hypotenuse.

Since the perimeter is 32, we have

$(1) \phantom{a} a+b+c=32$.

Since the area is 20 and the legs are $a$ and $b$, we have that

$(2) \phantom{a} \frac{a \cdot b}{2}=20$.

By the Pythagorean Theorem, we have that

$(3) \phantom{a} a^2+b^2=c^2$.

Since we want $c$, we will equations $1, 2, 3$ be in the form of $c.$

Equation 1 can be turned into

$(4) \phantom{a} a+b=32-c$.

Equation 2 can be simplified into

$(5) \phantom{a} ab=40.$

Equation 3 is already simplified.

Onto the calculating process.

Squaring the 1st equation we have

$(a+b+c)^2=32^2.$

Expanding and grouping, we have

$(a^2+b^2+c^2)+2(ab+ac+bc)=32^2.$

By equation 3 and substituting we get

$2(c^2+ab+ac+bc)=32^2.$

By equation 5 and substituting we get

$2(c^2+40+ac+bc)=32^2.$

Note that we can factor $c$ out in the inner expression, and we get

$2(c^2+40+c(a+b))=32^2.$

By equation 4 and substituting, we have

$2(c^2+40+c(32-c))=32^2.$

Expanding, we have

$2(c^2+40+32c-c^2)=32^2.$

Simplifying, we have

$2(40+32c)=32^2.$

Expanding again, we get

$80+64c=32^2.$

Dividing both sides by $16$ gets us

$4c+5=64$

Calculating gets us

$c=\boxed{\mathrm{(B) \frac{59}{4}}}$.

~mathboy282

Solution 8

This solution is very similar to Solution 1, except instead of subtracting $a+b$ from both sides in the first part, we subtract $\sqrt{a^2+b^2}$ from both sides, which gets us:

\begin{align*} a+b&=32-\sqrt{a^2+b^2}\\ (a+b)^2&=(32-\sqrt{a^2+b^2})^2\\ a^2+2ab+b^2&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}+a^2+b^2\\ 2ab&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}.\\ \end{align*}

We know that \[\frac{1}{2}ab=20\] so \[ab=40.\]

We can then substitute $40$ for $ab$ to get us:

\begin{align*} 80&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}\\ 2\cdot32\cdot\sqrt{a^2+b^2}&=32^2-80\\ 64\cdot\sqrt{a^2+b^2}&=32^2-80\\ \sqrt{a^2+b^2}&=\frac{32^2-80}{64}\\ \end{align*}

We know that $\sqrt{a^2+b^2}$ is the hypotenuse, so we only have to solve the right-hand side now.

\begin{align*} \sqrt{a^2+b^2}&=\frac{32^2-80}{64}\\ &=\frac{32^2}{64}-\frac{80}{64}\\ &=\frac{32\cdot32}{32\cdot2}-\frac{32\cdot\frac{5}{2}}{32\cdot2}\\ &=\frac{32}{2}-\frac{\frac{5}{2}}{2}\\ &=16-\frac{5}{4}\\ &=\frac{64}{4}-\frac{5}{4}\\ &=\boxed{\mathrm{(B)} \frac{59}{4}}\\ \end{align*}

~zlrara01

Video Solution

https://youtu.be/mm1fGEKhQSA

~savannahsolver

See Also

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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=2008_AMC_10A_Problems/Problem_18&oldid=176521"