Difference between revisions of "2003 AIME I Problems/Problem 12"
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== Solution == | == Solution == | ||
+ | |||
+ | ===Solution 1=== | ||
<center><asy> | <center><asy> | ||
real x = 1.60; /* arbitrary */ | real x = 1.60; /* arbitrary */ | ||
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Since <math>x \neq 280 - x</math>, <math>280 - 2x \neq 0</math> and thus | Since <math>x \neq 280 - x</math>, <math>280 - 2x \neq 0</math> and thus | ||
<math>360\cos A = 280</math> so <math>\cos A = \frac{7}{9} = 0.7777\ldots</math> and so <math>\lfloor 1000\cos A\rfloor = \boxed{777}</math>. | <math>360\cos A = 280</math> so <math>\cos A = \frac{7}{9} = 0.7777\ldots</math> and so <math>\lfloor 1000\cos A\rfloor = \boxed{777}</math>. | ||
+ | |||
+ | |||
+ | ===Solution 2=== | ||
+ | Notice that <math>AB = CD</math>, and <math>BD = DB</math>, and <math>\angle{DAB} \cong \angle{BCD}</math>, so we have side-side-angle matching on triangles <math>ABD</math> and <math>CDB</math>. Since the problem does not allow <math>\triangle{ABD} \cong \triangle{CDB}</math>, we know that <math>\angle{ADB}</math> is not a right angle, and there is a unique other triangle with the matching side-side-angle. | ||
+ | |||
+ | Extend <math>AD</math> to <math>C'</math> so that <math>\triangle{ABC'}</math> is isosceles with <math>AB = C'B</math>. Then notice that <math>\triangle{DC'B}</math> has matching side-side-angle, and yet <math>\triangle{ADB} \not\cong \triangle{C'DB}</math> because <math>\angle{ADB}</math> is not right. Therefore <math>\triangle{C'DB}</math> is the unique triangle mentioned above, so <math>\triangle{CDB}</math> is congruent, in some order of vertices, to <math>\triangle{C'DB}</math>. Since <math>\triangle{CDB} \cong \triangle{C'DB}</math> would imply <math>\triangle{CDB} = \triangle{C'DB}</math>, making quadrilateral <math>ABCD</math> degenerate, we must have <math>\triangle{CDB} \cong \triangle{C'BD}</math>. | ||
+ | |||
+ | Since the perimeter of <math>ABCD</math> is <math>640</math>, <math>AD + BC = 640 - 180 - 180 = 280</math>. Hence <math>280 = AD + BC = AD + DC'</math>. Drop the altitude of <math>\triangle{ABC'}</math> from <math>B</math> and call the foot <math>P</math>. Then right triangle trigonometry on <math>\triangle{APB}</math> shows that <math>\cos{A} = AP/AB = 140/180 = 7/9</math>, so <math>\lfloor 1000 \cos A \rfloor = \boxed{777}</math>. | ||
== See also == | == See also == |
Revision as of 22:27, 27 February 2009
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)
Let so
. By the Law of Cosines in
at angle
and in
at angle
,
Then
and grouping the
terms gives
.
Since ,
and thus
so
and so
.
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 |