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 "2015 AIME II Problems/Problem 7"

(→‎Solution #2)
m
 
(57 intermediate revisions by 12 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
Triangle <math>ABC</math> has side lengths <math>AB = 12</math>, <math>BC = 25</math>, and <math>CA = 17</math>. Rectangle <math>PQRS</math> has vertex <math>P</math> on <math>\overline{AB}</math>, vertex <math>Q</math> on <math>\overline{AC}</math>, and vertices <math>R</math> and <math>S</math> on <math>\overline{BC}</math>. In terms of the side length <math>PQ = w</math>, the area of <math>PQRS</math> can be expressed as the quadratic polynomial
+
Triangle <math>ABC</math> has side lengths <math>AB = 12</math>, <math>BC = 25</math>, and <math>CA = 17</math>. Rectangle <math>PQRS</math> has vertex <math>P</math> on <math>\overline{AB}</math>, vertex <math>Q</math> on <math>\overline{AC}</math>, and vertices <math>R</math> and <math>S</math> on <math>\overline{BC}</math>. In terms of the side length <math>PQ = \omega</math>, the area of <math>PQRS</math> can be expressed as the quadratic polynomial <cmath>Area(PQRS) = \alpha \omega - \beta \omega^2.</cmath>
−
 
 
−
Area(<math>PQRS</math>) = <math>\alpha w - \beta \cdot w^2</math>.
 
  
 
Then the coefficient <math>\beta = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
Then the coefficient <math>\beta = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
−
==Solution==
+
==Solution 1==
  
 
If <math>\omega = 25</math>,  the area of rectangle <math>PQRS</math> is <math>0</math>, so
 
If <math>\omega = 25</math>,  the area of rectangle <math>PQRS</math> is <math>0</math>, so
Line 13: Line 11:
 
<cmath>\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0</cmath>
 
<cmath>\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0</cmath>
  
−
and <math>\alpha = 25\beta</math>.  If <math>\omega = \frac{25}{2}</math>, we can reflect <math>APQ</math> over PQ, <math>PBS</math> over <math>PS</math>, and <math>QCR</math> over <math>QR</math> to completely cover rectangle <math>PQRS</math>, so the area of <math>PQRS</math> is half the area of the triangle.  Using Heron's formula, since <math>s = \frac{12 + 17 + 25}{2} = 27</math>,  
+
and <math>\alpha = 25\beta</math>.  If <math>\omega = \frac{25}{2}</math>, we can reflect <math>APQ</math> over <math>PQ</math>, <math>PBS</math> over <math>PS</math>, and <math>QCR</math> over <math>QR</math> to completely cover rectangle <math>PQRS</math>, so the area of <math>PQRS</math> is half the area of the triangle.  Using Heron's formula, since <math>s = \frac{12 + 17 + 25}{2} = 27</math>,  
  
 
<cmath> [ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90</cmath>
 
<cmath> [ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90</cmath>
Line 27: Line 25:
 
so the answer is <math>m + n = 36 + 125 = \boxed{161}</math>.
 
so the answer is <math>m + n = 36 + 125 = \boxed{161}</math>.
  
−
==Solution #2==
+
==Solution 2==
−
Diagram:\\
+
<asy>
−
(Diagram goes here)
+
unitsize(20);
−
\\
+
pair A,B,C,E,F,P,Q,R,S;
−
Similar triangles can also solve the problem.\\
+
A=(48/5,36/5);
−
First, solve for the area of the triangle. <math>[ABC] = 90</math>. This can be done by Heron's Formula or placing an <math>8-15-17</math> right triangle on <math>BC</math> and solving. (The <math>8</math> side would be collinear with line <math>AB</math>)\\
+
B=(0,0);
 +
C=(25,0);
 +
E=(48/5,0);
 +
F=(48/5,18/5);
 +
P=(24/5,18/5);
 +
Q=(173/10,18/5);
 +
S=(24/5,0);
 +
R=(173/10,0);
 +
draw(A--B--C--cycle);
 +
draw(P--Q);
 +
draw(Q--R);
 +
draw(R--S);
 +
draw(S--P);
 +
draw(A--E,dashed);
 +
label("$A$",A,N);
 +
label("$B$",B,SW);
 +
label("$C$",C,SE);
 +
label("$E$",E,SE);
 +
label("$F$",F,NE);
 +
label("$P$",P,NW);
 +
label("$Q$",Q,NE);
 +
label("$R$",R,SE);
 +
label("$S$",S,SW);
 +
draw(rightanglemark(B,E,A,12));
 +
dot(E);
 +
dot(F);
 +
</asy>
 +
 
 +
Similar triangles can also solve the problem.
 +
 
 +
First, solve for the area of the triangle. <math>[ABC] = 90</math>. This can be done by Heron's Formula or placing an <math>8-15-17</math> right triangle on <math>AC</math> and solving. (The <math>8</math> side would be collinear with line <math>AB</math>)
 +
 
 
After finding the area, solve for the altitude to <math>BC</math>. Let <math>E</math> be the intersection of the altitude from <math>A</math> and side <math>BC</math>. Then <math>AE = \frac{36}{5}</math>.  
 
After finding the area, solve for the altitude to <math>BC</math>. Let <math>E</math> be the intersection of the altitude from <math>A</math> and side <math>BC</math>. Then <math>AE = \frac{36}{5}</math>.  
−
Solving for <math>BE</math> using the Pythagorean Formula, we get <math>BE = \frac{48}{5}</math>. We then know that <math>CE = \frac{77}{5}</math>.\\
+
Solving for <math>BE</math> using the Pythagorean Formula, we get <math>BE = \frac{48}{5}</math>. We then know that <math>CE = \frac{77}{5}</math>.
−
Now consider the rectangle <math>PQRS</math>. Since <math>SR</math> is collinear with <math>BC</math> and parallel to <math>PQ</math>, <math>PQ</math> is parallel to <math>BC</math> meaning <math>\Delta APQ</math> is similar to <math>\Delta ABC</math>. \\
+
 
−
Let <math>F</math> be the intersection between <math>AE</math> and <math>PQ</math>. By the similar triangles, we know that <math>\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}</math>. Since <math>PF+FQ=PQ=\omega</math>. We can solve for <math>PF</math> and <math>FQ</math> in terms of <math>\omega</math>. We get that <math>PF=\frac{48}{125} \omega</math> and <math>FQ=\frac{77}{125} \omega</math>.\\
+
Now consider the rectangle <math>PQRS</math>. Since <math>SR</math> is collinear with <math>BC</math> and parallel to <math>PQ</math>, <math>PQ</math> is parallel to <math>BC</math> meaning <math>\Delta APQ</math> is similar to <math>\Delta ABC</math>.  
−
Let's work with <math>PF</math>. We know that <math>PQ</math> is parallel to <math>BC</math> so <math>\Delta APF</math> is similar to <math>\Delta ABE</math>. We can set up the proportion:\\
+
 
 +
Let <math>F</math> be the intersection between <math>AE</math> and <math>PQ</math>. By the similar triangles, we know that <math>\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}</math>. Since <math>PF+FQ=PQ=\omega</math>. We can solve for <math>PF</math> and <math>FQ</math> in terms of <math>\omega</math>. We get that <math>PF=\frac{48}{125} \omega</math> and <math>FQ=\frac{77}{125} \omega</math>.
 +
 
 +
Let's work with <math>PF</math>. We know that <math>PQ</math> is parallel to <math>BC</math> so <math>\Delta APF</math> is similar to <math>\Delta ABE</math>. We can set up the proportion:
 +
 
 
<math>\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}</math>. Solving for <math>AF</math>, <math>AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega</math>.  
 
<math>\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}</math>. Solving for <math>AF</math>, <math>AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega</math>.  
−
We can solve for <math>PS</math> then since we know that <math>PS=FE</math> and <math>FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega</math>.\\
+
 
−
Therefore, <math>[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2</math>.\\
+
We can solve for <math>PS</math> then since we know that <math>PS=FE</math> and <math>FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega</math>.
 +
 
 +
Therefore, <math>[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2</math>.
 +
 
 
This means that <math>\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}</math>.
 
This means that <math>\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}</math>.
 +
 +
==Solution 3==
 +
Heron's Formula gives <math>[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90,</math> so the altitude from <math>A</math> to <math>BC</math> has length <math>\dfrac{2[ABC]}{BC} = \dfrac{36}{5}.</math>
 +
 +
Now, draw a parallel to <math>AB</math> from <math>Q</math>, intersecting <math>BC</math> at <math>T</math>. Then <math>BT = w</math> in parallelogram <math>QPBT</math>, and so <math>CT = 25 - w</math>. Clearly, <math>CQT</math> and <math>CAB</math> are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so
 +
<cmath>\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.</cmath>
 +
Solving gives <math>[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} \cdot w = \dfrac{36w}{5} - \dfrac{36w^2}{125}</math>, so the answer is <math>36 + 125 = 161</math>.
 +
 +
==Solution 4==
 +
Using the diagram from Solution 2 above, label <math>AF</math> to be <math>h</math>. Through Heron's formula, the area of <math>\triangle ABC</math> turns out to be <math>90</math>, so using <math>AE</math> as the height and <math>BC</math> as the base yields <math>AE=\frac{36}{5}</math>. Now, through the use of similarity between <math>\triangle APQ</math> and <math>\triangle ABC</math>, you find <math>\frac{w}{25}=\frac{h}{36/5}</math>. Thus, <math>h=\frac{36w}{125}</math>. To find the height of the rectangle, subtract <math>h</math> from <math>\frac{36}{5}</math> to get <math>\left(\frac{36}{5}-\frac{36w}{125}\right)</math>, and multiply this by the other given side <math>w</math> to get <math>\frac{36w}{5}-\frac{36w^2}{125}</math> for the area of the rectangle. Finally, <math>36+125=\boxed{161}</math>.
 +
 +
==Solution 5==
 +
Using the diagram as shown in Solution 2, let <math>AE=h</math> and <math>AP=L</math>
 +
Now, by Heron's formula, we find that the <math>[ABC]=90</math>. Hence, <math>h=\frac{36}{5}</math>
 +
 +
Now, we see that <math>\sin{B}=\frac{PS}{12-L}\implies PS=\sin{B}(12-L)</math>
 +
We easily find that <math>\sin{B}=\frac{3}{5}</math>
 +
 +
Hence, <math>PS=\frac{3}{5}(12-L)</math>
 +
 +
Now, we see that <math>[PQRS]=\frac{3}{5}(12-L)(w)</math>
 +
 +
Now, it is obvious that we want to find <math>L</math> in terms of <math>W</math>.
 +
 +
Looking at the diagram, we see that because <math>PQRS</math> is a rectangle, <math>\triangle{APQ}\sim{\triangle{ABC}}</math>
 +
 +
Hence.. we can now set up similar triangles.
 +
 +
We have that <math>\frac{AP}{AB}=\frac{PQ}{BC}\implies \frac{L}{12}=\frac{W}{25}\implies 25L=12W\implies L=\frac{12W}{25}</math>.
 +
 +
Plugging back in..
 +
 +
 +
<math>[PQRS]=\frac{3w}{5}(12-(\frac{12W}{25}))\implies \frac{3w}{5}(\frac{300-12W}{25})\implies \frac{900W-36W^2}{125}</math>
 +
 +
Simplifying, we get <math>\frac{36W}{5}-\frac{36W^2}{125}</math>
 +
 +
Hence, <math>125+36=\boxed{161}</math>
 +
 +
==Solution 6==
 +
Proceed as in solution 1. When <math>\omega</math> is equal to zero, <math>\alpha - \beta\omega=\alpha</math> is equal to the altitude. This means that <math>25\beta</math> is equal to <math>\frac{36}{5}</math>, so <math>\beta = \frac{36}{125}</math>, yielding <math>\boxed{161}</math>.
 +
 +
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=9re2qLzOKWk&t=554s
 +
 +
~MathProblemSolvingSkills.com
 +
 +
  
 
==See also==
 
==See also==
 
{{AIME box|year=2015|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2015|n=II|num-b=6|num-a=8}}
 +
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:08, 14 February 2023

Problem

Triangle $ABC$ has side lengths $AB = 12$, $BC = 25$, and $CA = 17$. Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$, vertex $Q$ on $\overline{AC}$, and vertices $R$ and $S$ on $\overline{BC}$. In terms of the side length $PQ = \omega$, the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]

Then the coefficient $\beta = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

If $\omega = 25$, the area of rectangle $PQRS$ is $0$, so

\[\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0\]

and $\alpha = 25\beta$. If $\omega = \frac{25}{2}$, we can reflect $APQ$ over $PQ$, $PBS$ over $PS$, and $QCR$ over $QR$ to completely cover rectangle $PQRS$, so the area of $PQRS$ is half the area of the triangle. Using Heron's formula, since $s = \frac{12 + 17 + 25}{2} = 27$,

\[[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90\]

so

\[45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}\]

and

\[\beta = \frac{180}{625} = \frac{36}{125}\]

so the answer is $m + n = 36 + 125 = \boxed{161}$.

Solution 2

[asy] unitsize(20); pair A,B,C,E,F,P,Q,R,S; A=(48/5,36/5); B=(0,0); C=(25,0); E=(48/5,0); F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$E$",E,SE); label("$F$",F,NE); label("$P$",P,NW); label("$Q$",Q,NE); label("$R$",R,SE); label("$S$",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); [/asy]

Similar triangles can also solve the problem.

First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$)

After finding the area, solve for the altitude to $BC$. Let $E$ be the intersection of the altitude from $A$ and side $BC$. Then $AE = \frac{36}{5}$. Solving for $BE$ using the Pythagorean Formula, we get $BE = \frac{48}{5}$. We then know that $CE = \frac{77}{5}$.

Now consider the rectangle $PQRS$. Since $SR$ is collinear with $BC$ and parallel to $PQ$, $PQ$ is parallel to $BC$ meaning $\Delta APQ$ is similar to $\Delta ABC$.

Let $F$ be the intersection between $AE$ and $PQ$. By the similar triangles, we know that $\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}$. Since $PF+FQ=PQ=\omega$. We can solve for $PF$ and $FQ$ in terms of $\omega$. We get that $PF=\frac{48}{125} \omega$ and $FQ=\frac{77}{125} \omega$.

Let's work with $PF$. We know that $PQ$ is parallel to $BC$ so $\Delta APF$ is similar to $\Delta ABE$. We can set up the proportion:

$\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}$. Solving for $AF$, $AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega$.

We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega$.

Therefore, $[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2$.

This means that $\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}$.

Solution 3

Heron's Formula gives $[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90,$ so the altitude from $A$ to $BC$ has length $\dfrac{2[ABC]}{BC} = \dfrac{36}{5}.$

Now, draw a parallel to $AB$ from $Q$, intersecting $BC$ at $T$. Then $BT = w$ in parallelogram $QPBT$, and so $CT = 25 - w$. Clearly, $CQT$ and $CAB$ are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so \[\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.\] Solving gives $[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} \cdot w = \dfrac{36w}{5} - \dfrac{36w^2}{125}$, so the answer is $36 + 125 = 161$.

Solution 4

Using the diagram from Solution 2 above, label $AF$ to be $h$. Through Heron's formula, the area of $\triangle ABC$ turns out to be $90$, so using $AE$ as the height and $BC$ as the base yields $AE=\frac{36}{5}$. Now, through the use of similarity between $\triangle APQ$ and $\triangle ABC$, you find $\frac{w}{25}=\frac{h}{36/5}$. Thus, $h=\frac{36w}{125}$. To find the height of the rectangle, subtract $h$ from $\frac{36}{5}$ to get $\left(\frac{36}{5}-\frac{36w}{125}\right)$, and multiply this by the other given side $w$ to get $\frac{36w}{5}-\frac{36w^2}{125}$ for the area of the rectangle. Finally, $36+125=\boxed{161}$.

Solution 5

Using the diagram as shown in Solution 2, let $AE=h$ and $AP=L$ Now, by Heron's formula, we find that the $[ABC]=90$. Hence, $h=\frac{36}{5}$

Now, we see that $\sin{B}=\frac{PS}{12-L}\implies PS=\sin{B}(12-L)$ We easily find that $\sin{B}=\frac{3}{5}$

Hence, $PS=\frac{3}{5}(12-L)$

Now, we see that $[PQRS]=\frac{3}{5}(12-L)(w)$

Now, it is obvious that we want to find $L$ in terms of $W$.

Looking at the diagram, we see that because $PQRS$ is a rectangle, $\triangle{APQ}\sim{\triangle{ABC}}$

Hence.. we can now set up similar triangles.

We have that $\frac{AP}{AB}=\frac{PQ}{BC}\implies \frac{L}{12}=\frac{W}{25}\implies 25L=12W\implies L=\frac{12W}{25}$.

Plugging back in..


$[PQRS]=\frac{3w}{5}(12-(\frac{12W}{25}))\implies \frac{3w}{5}(\frac{300-12W}{25})\implies \frac{900W-36W^2}{125}$

Simplifying, we get $\frac{36W}{5}-\frac{36W^2}{125}$

Hence, $125+36=\boxed{161}$

Solution 6

Proceed as in solution 1. When $\omega$ is equal to zero, $\alpha - \beta\omega=\alpha$ is equal to the altitude. This means that $25\beta$ is equal to $\frac{36}{5}$, so $\beta = \frac{36}{125}$, yielding $\boxed{161}$.


Video Solution

https://www.youtube.com/watch?v=9re2qLzOKWk&t=554s

~MathProblemSolvingSkills.com


See also

2015 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2015_AIME_II_Problems/Problem_7&oldid=189616"