Difference between revisions of "2003 AIME I Problems/Problem 12"
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<cmath>180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A </cmath> | <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) | + | <cmath>(AD^2 - BC^2) = 360(AD - BC) \cos A</cmath> |
− | <cmath>(AD - BC)(AD + BC - | + | <cmath>(AD - BC)(AD + BC) = 360(AD - BC) \cos A</cmath> |
− | < | + | <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=== | ===Solution 2=== |
Revision as of 13:26, 3 July 2015
Contents
[hide]Problem
In convex quadrilateral and
The perimeter of
is 640. Find
(The notation
means the greatest integer that is less than or equal to
)
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]](http://latex.artofproblemsolving.com/0/6/0/060d4d918b4f89b6f156f9af44aea8da1238fcdd.png)
By the Law of Cosines on at angle
and on
at angle
(note
),
We know that
.
.
Solution 2
Notice that , and
, and
, so we have side-side-angle matching on triangles
and
. Since the problem does not allow
, we know that
is not a right angle, and there is a unique other triangle with the matching side-side-angle.
Extend to
so that
is isosceles with
. Then notice that
has matching side-side-angle, and yet
because
is not right. Therefore
is the unique triangle mentioned above, so
is congruent, in some order of vertices, to
. Since
would imply
, making quadrilateral
degenerate, we must have
.
Since the perimeter of is
,
. Hence
. Drop the altitude of
from
and call the foot
. Then right triangle trigonometry on
shows that
, so
.
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.