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2003 AIME I Problems/Problem 12

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Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

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

real x = 1.60; /* arbitrary */pointpen = black; pathpen = black+linewidth(0.7); size(180);real BD = x*x + 1.80*1.80 - 2 * 1.8...

Let $AD = x$ so $BC = 640 - 360 - x = 280 - x$ . By the Law of Cosines in $\triangle ABD$ at angle $A$ and in $\triangle BCD$ at angle $C$ , $180^2 + x^2 - 2\cdot180 \cdot x \cdot \cos A = BD^2 = 180^2 + (280 - x)^2 - 2\cdot180\cdot(280 - x) \cdot \cos A.$ Then $x^2 - 360x\cos A = (280 -x)^2 -360(280 - x)\cos A$ and grouping the $\cos A$ terms gives $360(280 - 2x)\cos A = 280(280 - 2x)$ .

Since $x \neq 280 - x$ , $280 - 2x \neq 0$ and thus $360\cos A = 280$ so $\cos A = \frac{7}{9} = 0.7777\ldots$ and so $\lfloor 1000\cos A\rfloor = \boxed{777}$ .

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.

Extend $AD$ to $C'$ so that $\triangle{ABC'}$ is isosceles with $AB = C'B$ . Then notice that $\triangle{DC'B}$ has matching side-side-angle, and yet $\triangle{ADB} \not\cong \triangle{C'DB}$ because $\angle{ADB}$ is not right. Therefore $\triangle{C'DB}$ is the unique triangle mentioned above, so $\triangle{CDB}$ is congruent, in some order of vertices, to $\triangle{C'DB}$ . Since $\triangle{CDB} \cong \triangle{C'DB}$ would imply $\triangle{CDB} = \triangle{C'DB}$ , making quadrilateral $ABCD$ degenerate, we must have $\triangle{CDB} \cong \triangle{C'BD}$ .

Since the perimeter of $ABCD$ is $640$ , $AD + BC = 640 - 180 - 180 = 280$ . Hence $280 = AD + BC = AD + DC'$ . Drop the altitude of $\triangle{ABC'}$ from $B$ and call the foot $P$ . Then right triangle trigonometry on $\triangle{APB}$ shows that $\cos{A} = AP/AB = 140/180 = 7/9$ , so $\lfloor 1000 \cos A \rfloor = \boxed{777}$ .

See also

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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2003_AIME_I_Problems/Problem_12"

Category: Intermediate Geometry Problems

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