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Difference between revisions of "2004 AIME II Problems/Problem 7"

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It is clear from the problem setup that <math>0<\theta<\frac\pi2</math>, so the correct value is <math>\tan(\theta)=\frac53</math>. Next, extend rays <math>\overrightarrow{BC}</math> and <math>\overrightarrow{EF}</math> to intersect at <math>C'</math>. Then <math>\tan(\theta)=\frac{BC'}{17}=\frac53</math>, so <math>BC'=\frac{85}{3}</math>. By similar triangles, <math>CC'=\frac{3}{17}BC'=\frac{15}{3}</math>, so <math>BC=\frac{70}{3}</math>. The perimeter is <math>\frac{140}{3}+50=\frac{290}{3}\Longrightarrow \boxed{293}</math>
 
It is clear from the problem setup that <math>0<\theta<\frac\pi2</math>, so the correct value is <math>\tan(\theta)=\frac53</math>. Next, extend rays <math>\overrightarrow{BC}</math> and <math>\overrightarrow{EF}</math> to intersect at <math>C'</math>. Then <math>\tan(\theta)=\frac{BC'}{17}=\frac53</math>, so <math>BC'=\frac{85}{3}</math>. By similar triangles, <math>CC'=\frac{3}{17}BC'=\frac{15}{3}</math>, so <math>BC=\frac{70}{3}</math>. The perimeter is <math>\frac{140}{3}+50=\frac{290}{3}\Longrightarrow \boxed{293}</math>
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 +
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An even faster way to finish is, to draw a line segment <math>FF'</math> where <math>F'</math> is a point on <math>EB</math> such that <math>FF'</math> is perpendicular to <math>EB</math>. This makes right triangle <math>FF'E</math>, Also, note that <math>F'B</math> has length of <math>3</math> (draw the diagram out, and note the <math>F'B =FC</math>). From here, through <math>\tan \theta = \frac{5}{3}</math>, we can note that <math>\frac{FF'}{EF'} = \frac{5}{3} \implies \frac{FF'}{14} = \frac{5}{3} \implies FF' = \frac{70}{3}</math>. <math>FF'</math> is parallel and congrurent to <math>CB</math> and <math>AD</math>, and hence we can use this to calculate the perimeter. The perimeter is simply <math>\frac{70}{3} + \frac{70}{3} + 25 + 25 = \frac{290}{3} \Longrightarrow \boxed{293}</math>
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== See also ==
 
== See also ==
 
{{AIME box|year=2004|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2004|n=II|num-b=6|num-a=8}}

Revision as of 11:14, 26 November 2017

Problem

$ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

Solution 1 (synthetic)

[asy] pointpen = black; pathpen = black +linewidth(0.7); pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,W)--MP("F",F,(1,0))); D(B--G); D(E--MP("B'",G)--F--B,dashed); MP("8",(A+E)/2,W);MP("17",(B+E)/2,W);MP("22",(D+F)/2,(1,0)); [/asy]

Since $EF$ is the perpendicular bisector of $\overline{BB'}$, it follows that $BE = B'E$ (by SAS). By the Pythagorean Theorem, we have $AB' = 15$. Similarly, from $BF = B'F$, we have \begin{align*} BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\ BC &= \frac{70}{3} \end{align*} Thus the perimeter of $ABCD$ is $2\left(25 + \frac{70}{3}\right) = \frac{290}{3}$, and the answer is $m+n=\boxed{293}$.

Solution 2 (analytic)

Let $A = (0,0), B=(0,25)$, so $E = (0,8)$ and $F = (l,22)$, and let $l = AD$ be the length of the rectangle. The slope of $EF$ is $\frac{14}{l}$ and so the equation of $EF$ is $y -8 = \frac{14}{l}x$. We know that $EF$ is perpendicular to and bisects $BB'$. The slope of $BB'$ is thus $\frac{-l}{14}$, and so the equation of $BB'$ is $y -25 = \frac{-l}{14}x$. Let the point of intersection of $EF, BB'$ be $G$. Then the y-coordinate of $G$ is $\frac{25}{2}$, so \begin{align*} \frac{14}{l}x &= y-8 = \frac{9}{2}\\ \frac{-l}{14}x &= y-25 = -\frac{25}{2}\\ \end{align*} Dividing the two equations yields

$l^2 = \frac{25 \cdot 14^2}{9} \Longrightarrow l = \frac{70}{3}$

The answer is $\boxed{293}$ as above.

Solution 3 (Coordinate Bashing)

Firstly, observe that if we are given that $AE=8$ and $BE=17$, the length of the triangle is given and the height depends solely on the length of $CF$. Let Point $A = (0,0)$. Since $AE=8$, point E is at (8,0). Next, point $B$ is at $(25,0)$ since $BE=17$ and point $B'$ is at $(0,-15)$ since $BE=AE$ by symmetry. Draw line segment $BB'$. Notice that this is perpendicular to $EF$ by symmetry. Next, find the slope of EB, which is $\frac{15}{25}=\frac{3}{5}$. Then, the slope of $EF$ is -$\frac{5}{3}$.

Line EF can be written as y=$-\frac{5}{3}x+b$. Plug in the point $(8,0)$, and we get the equation of EF to be y=$_\frac{5}{3}x+\frac{40}{3}$. Since the length of $AB$=25, a point on line $EF$ lies on $DC$ when $x=25-3=22$. Plug in $x=22$ into our equation to get $y=-\frac{70}{3}$. $|y|=BC=\frac{70}{3}$. Therefore, our answer is $2(AB+BC)=2\left(25+\frac{70}{3}\right)=2\left(\frac{145}{3}\right)=\frac{290}{3}= \boxed{293}$.

Solution 4 (Trig)

Firstly, note that $B'E=BE=17$, so $AB'=\sqrt{17^2-8^2}=15$. Then let $\angle BEF=\angle B'EF=\theta$, so $\angle B'EA = \pi-2\theta$. Then $\tan(\pi-2\theta)=\frac{15}{8}$, or

\[\frac{2\tan(\theta)}{\tan^2(\theta)-1}=\frac{15}{8}\] using supplementary and double angle identities. Multiplying though and factoring yields

\[(3\tan(\theta)-5)(5\tan(\theta)+3)=0\]

It is clear from the problem setup that $0<\theta<\frac\pi2$, so the correct value is $\tan(\theta)=\frac53$. Next, extend rays $\overrightarrow{BC}$ and $\overrightarrow{EF}$ to intersect at $C'$. Then $\tan(\theta)=\frac{BC'}{17}=\frac53$, so $BC'=\frac{85}{3}$. By similar triangles, $CC'=\frac{3}{17}BC'=\frac{15}{3}$, so $BC=\frac{70}{3}$. The perimeter is $\frac{140}{3}+50=\frac{290}{3}\Longrightarrow \boxed{293}$


An even faster way to finish is, to draw a line segment $FF'$ where $F'$ is a point on $EB$ such that $FF'$ is perpendicular to $EB$. This makes right triangle $FF'E$, Also, note that $F'B$ has length of $3$ (draw the diagram out, and note the $F'B =FC$). From here, through $\tan \theta = \frac{5}{3}$, we can note that $\frac{FF'}{EF'} = \frac{5}{3} \implies \frac{FF'}{14} = \frac{5}{3} \implies FF' = \frac{70}{3}$. $FF'$ is parallel and congrurent to $CB$ and $AD$, and hence we can use this to calculate the perimeter. The perimeter is simply $\frac{70}{3} + \frac{70}{3} + 25 + 25 = \frac{290}{3} \Longrightarrow \boxed{293}$

See also

2004 AIME II (Problems • Answer Key • Resources)
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Problem 8
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