AIME 2004 II · 第 7 题

Solutions

Solution 1 (Synthetic)

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

$$\begin{aligned} BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\ BC &= \frac{70}{3} \end{aligned}$$ 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

$$\begin{aligned} \frac{14}{l}x &= y-8 = \frac{9}{2}\\ \frac{-l}{14}x &= y-25 = -\frac{25}{2}\\ \end{aligned}$$ 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