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

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The perimeter is <math>\frac{140}{3}+50=\frac{290}{3},</math> so our answer is <math>\boxed{293}</math>.
 
The perimeter is <math>\frac{140}{3}+50=\frac{290}{3},</math> so our answer is <math>\boxed{293}</math>.
  
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===Solution 7===
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===Solution 7 (Similar to solution 5, more in depth)===
 
Let the endpoint of the intersection of the fold near <math>F</math> be <math>G</math>. Since trapezoid <math>BCFE</math> is folded, it is congruent to trapezoid <math>B'C'FE</math>. Therefore, <math>BE=B'E=17</math>. Since <math>\triangle AB'E</math> is a right triangle, <math>AB'=15</math> from the pythagorean theorem. From here, we can see that triangles <math>\triangle AEB \sim \triangle DGB' \sim \triangle C'GF</math> by AA similarity. From here, we find <math>BC</math> from a lot of similarities. Let <math>BC=x</math>.
 
Let the endpoint of the intersection of the fold near <math>F</math> be <math>G</math>. Since trapezoid <math>BCFE</math> is folded, it is congruent to trapezoid <math>B'C'FE</math>. Therefore, <math>BE=B'E=17</math>. Since <math>\triangle AB'E</math> is a right triangle, <math>AB'=15</math> from the pythagorean theorem. From here, we can see that triangles <math>\triangle AEB \sim \triangle DGB' \sim \triangle C'GF</math> by AA similarity. From here, we find <math>BC</math> from a lot of similarities. Let <math>BC=x</math>.
 
Since <math>\triangle ABE' \sim \triangle DGB'</math>:
 
Since <math>\triangle ABE' \sim \triangle DGB'</math>:
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Finally, our answer is <math>2(\frac {70}{3}) + 2(25)=\frac {290}{3}</math>, which is <math>290+3=\boxed{293}</math>.
 
Finally, our answer is <math>2(\frac {70}{3}) + 2(25)=\frac {290}{3}</math>, which is <math>290+3=\boxed{293}</math>.
  
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~ Wesserwessey7254
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<math>\frac {AB}{CD} = \frac {3}{4}</math>.
 
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- Wesserwessey7254
 
  
 
== See also ==
 
== See also ==

Revision as of 04:33, 30 October 2023

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 $m+n.$

[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy]

Solutions

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 our 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 $\frac{70}{3} + \frac{70}{3} + 25 + 25 = \frac{290}{3} \Longrightarrow \boxed{293}$

Solution 5 (Fast, Pythagorean)

Use the prepared diagram for this solution.

Call the intersection of DF and B'C' G. AB'E is an 8-15-17 right triangle, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculated to be 25/3. Add all the sides together to get $\boxed{293}$.

-jackshi2006

Solution 6(fast as wind[rufeng])

Call the intersection of $B'C'$, $BC$, and $EF$ $G$. Since $FCBE$ and $FC'B'E$ are congruent, we know that the three lines intersect. We already know $AB$ so we just need to find $CB$, call it $x$. Drop an altitude from $F$ to $AB$ and call it $H$. $EH=EB-FC=14$. Using Pythagorean Theorem, we have $EF=\sqrt{x^2+14^2}$. Triangles $EFH$ and $EGB$ are similar (AA), so we get \[\frac{HF}{BG}=\frac{EH}{EB}\] \[\frac{x}{x+GC}=\frac{14}{17}\] Simplify and we get $GC=\frac{3x}{14}$.

We find the area of $FCBE$ by using the fact that it is a trapezoid. $[FCBE]=\frac{(3+17)x}{2}=10x$

A different way to find the area: $[FCBE]=\frac{1}{2} EG\cdot($height of $EGB$ with $EG$ as base$)-[FGC]$

Since $GBE$ and $G'B'E$ are congruent(SAS), their height from $EG$ is the same. $B'B=\sqrt{AB'^2+AB^2}=5\sqrt{34}$. $EG=\sqrt{EB^2+BG^2}=\sqrt{(\frac{17x}{14})^2+17^2}=17\sqrt{\frac{x^2}{196}+1}$

\[[FCBE]=\frac{1}{2} \cdot 17 \cdot \sqrt{\frac{x^2}{196}+1} \cdot \frac{5\sqrt{34}}{2}-\frac{9x}{28}\] \[280x+9x=7\cdot 5 \cdot \sqrt{34} \cdot 17 \cdot \sqrt{\frac{x^2}{196}+1}\] \[17^4 x^2=49 \cdot 25 \cdot 34 \cdot 17^2 \cdot (\frac{x^2}{196}+1)\] \[17x^2=\frac{25}{2}x^2+2450\] \[x=\frac{70}{3}\]

The perimeter is $\frac{140}{3}+50=\frac{290}{3},$ so our answer is $\boxed{293}$.

Solution 7 (Similar to solution 5, more in depth)

Let the endpoint of the intersection of the fold near $F$ be $G$. Since trapezoid $BCFE$ is folded, it is congruent to trapezoid $B'C'FE$. Therefore, $BE=B'E=17$. Since $\triangle AB'E$ is a right triangle, $AB'=15$ from the pythagorean theorem. From here, we can see that triangles $\triangle AEB \sim \triangle DGB' \sim \triangle C'GF$ by AA similarity. From here, we find $BC$ from a lot of similarities. Let $BC=x$. Since $\triangle ABE' \sim \triangle DGB'$:

\[\frac {AE}{AB'} = \frac{DB}{DG}\]

\[\frac {8}{15} = \frac {x-15}{DG}\]

\[DG = \frac {15(x-15)}{8}\]

\[GF = DC-DG-FC\]

\[GF = \frac{-15x+401}{8}\]

Since $\triangle ABE' \sim \triangle C'GF'$,

\[\frac {AE}{B'E} = \frac {C'F}{GF}\]

\[\frac {8}{17} = \frac{3}{\frac {-15x+401}{8}}\]

from which we get $x= \frac {70}{3}$.

Finally, our answer is $2(\frac {70}{3}) + 2(25)=\frac {290}{3}$, which is $290+3=\boxed{293}$.

~ Wesserwessey7254

See also

2004 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

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