Difference between revisions of "2015 AIME II Problems/Problem 7"

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

@@ Line 1: / Line 1: @@
 ==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>.
-==Solution==
+==Solution 1==
 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>
-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>
@@ Line 27: / Line 25: @@
 so the answer is <math>m + n = 36 + 125 = \boxed{161}</math>.
-==Solution #2==
+==Solution 2==
-Diagram:
 <asy>
-unitsize(35);
+unitsize(20);
 pair A,B,C,E,F,P,Q,R,S;
-B=(0,0); S=(24/5,0); E = (48/5,0); R=(173/10,0);C=(25,0);A=(48/5,36/5);F=(48/5,18/5);P=(24/5,18/5);Q=(173/10,18/5);
+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. <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>)
+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>.
@@ Line 56: / Line 79: @@
 This means that <math>\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}</math>.
-- solution by abvenkgoo
+==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==
 {{AIME box|year=2015|n=II|num-b=6|num-a=8}}
+[[Category: Intermediate Geometry Problems]]
 {{MAA Notice}}

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