Difference between revisions of "2003 AIME I Problems/Problem 12"

Revision as of 13:26, 3 July 2015

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 $AD + BC = 640 - 360 = 280$ . $\cos A = \dfrac{240}{360} = \dfrac{7}{9} = 0.777 \ldots$

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

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.

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

@@ Line 18: / Line 18: @@
 <cmath>180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A </cmath>
-<cmath>(AD^2 - BC^2) - 360(AD - BC) \cos A = 0 </cmath>
+<cmath>(AD^2 - BC^2) = 360(AD - BC) \cos A</cmath>
-<cmath>(AD - BC)(AD + BC - 360 \cos A ) = 0 </cmath>
+<cmath>(AD - BC)(AD + BC) = 360(AD - BC) \cos A</cmath>
-<math>AD - BC \neq 0</math>, so <math>AD + BC - 360 \cos A = 0</math>. We know that <math>AD + BC = 640 - 360 = 280</math>, so <math>280 - 360 \cos A = 0 \Rightarrow \cos A = \dfrac{7}{9} = 0.777 \ldots</math>, and <math>\lfloor 1000 \cos  A \rfloor = \boxed{777}</math>.
+<cmath>(AD + BC) = 360 \cos A</cmath>
+We know that <math>AD + BC = 640 - 360 = 280</math>. <math>\cos A = \dfrac{240}{360} =  \dfrac{7}{9} = 0.777 \ldots</math>
+<math>\lfloor 1000 \cos  A \rfloor = \boxed{777}</math>.
 ===Solution 2===

2003 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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