2003 AIME I Problems/Problem 12

Problem

In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$ . Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$ )

Solution

Solution 1

$[asy] real x = 1.60; /* arbitrary */ pointpen = black; pathpen = black+linewidth(0.7); size(180); real BD = x*x + 1.80*1.80 - 2 * 1.80 * x * 7 / 9; pair A=(0,0),B=(1.8,0),D=IP(CR(A,x),CR(B,BD)),C=OP(CR(D,1.8),CR(B,2.80 - x)); D(MP("A",A)--MP("B",B)--MP("C",C)--MP("D",D,N)--B--A--D); MP("180",(A+B)/2); MP("180",(C+D)/2,NE); D(anglemark(B,A,D)); D(anglemark(D,C,B)); [/asy]$

By the Law of Cosines on $\triangle ABD$ at angle $A$ and on $\triangle BCD$ at angle $C$ (note $\angle C = \angle A$ ),

$180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A$ $(AD^2 - BC^2) = 360(AD - BC) \cos A$ $(AD - BC)(AD + BC) = 360(AD - BC) \cos A$ $(AD + BC) = 360 \cos A$ We know that $AD + BC = 640 - 360 = 280$ . $\cos A = \dfrac{280}{360} = \dfrac{7}{9} = 0.777 \ldots$

$\lfloor 1000 \cos A \rfloor = \boxed{777}$ .

Solution 2

Notice that $AB = CD$ , and $BD = DB$ , and $\angle{DAB} \cong \angle{BCD}$ , so we have side-side-angle matching on triangles $ABD$ and $CDB$ . Since the problem does not allow $\triangle{ABD} \cong \triangle{CDB}$ , we know that $\angle{ADB}$ is not a right angle, and there is a unique other triangle with the matching side-side-angle.

$[asy] pair A,B,C,D,F; A=(0,0);B=(5,4.04061);C=(6,0);D=(4,0);F=(5,0); draw(B--A--C--B--D); label("A",A,SW,red);label("E",F,S);label("B",B,NW,red);label("D",D,S,red);label("B",C,SE,blue);label("C",A,NW,blue);label("D",B,NE,blue);label("$\theta$",A,(4.5,1.5)); draw(anglemark(D,A,B,20));label("180",(A+B)/2,NW); draw(B--F,dashed);draw(rightanglemark(B,F,A));draw(D--B--C,dashed); add(pathticks(B--C,1,0.5,1,10));add(pathticks(B--D,1,0.5,1,10)); [/asy]$

Overlay the triangles $\triangle BDA$ and $\triangle BDC$ on each other as in the diagram above (where the red labels correspond to $\triangle BDA$ and the blue labels correspond to $\triangle BDC$ ). Here we assume without loss of generality that $BC>AD$ . Furthermore, let $\theta$ be the angle $A$ referenced in the problem; we need to find $\lfloor 1000\cos\theta\rfloor$ .

Since the perimeter of $ABCD$ is $640$ , we have $AD+BC=640-360=280$ . Thus let $BC=a$ and $AD=280-a$ for some positive real number $a$ . But the sides that correspond to $\overline{BD}$ above are congruent, so we can drop a perpendicular from the topmost point to a point $E$ , where the base of the isosceles triangle is bisected. Notice that the base of the isosceles triangle has length $a-(280-a)=2a-280$ , so in the diagram above, $AE=280-a+\frac{1}{2}(2a-280)=280-a+a-140=140$ .

Looking above at the right triangle containing $\theta$ , we see that $\cos\theta=\frac{AE}{180}=\frac{140}{180}=\frac{7}{9}$ . Hence $\left\lfloor 1000\cdot\frac{7}{9}\right\rfloor=\boxed{777}$ is our answer.

~ eevee9406

Solution 3

Start the same as solution 1. We get $180^2+a^2-360a \cos A = 180^2+b^2-360b \cos A \Rightarrow a^2-360a \cos A = b^2-360b \cos A,$ where $a$ is the length of $BC$ and $b$ is the length of $AD$ . Let the common value of $a^2-360a \cos A$ and $b^2-360b \cos A$ be $c$ . Then, the quadratic in $x$ $x^2-360 \cos A \cdot x - c = 0$ has solutions $a$ and $b$ . Therefore, by Vieta's, $a+b = 360 \cos A$ . However, we know that the perimeter of $ABCD$ is $640$ , so $a+b+180+180=640$ , so $a+b=280$ . Therefore, $360 \cos A = 280 \Rightarrow \cos A = \frac{280}{360} \Rightarrow \cos A = \frac{7}{9} \Rightarrow \lfloor 1000 \cos A \rfloor = \boxed{777}.$

2003 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search