Difference between revisions of "2003 AIME I Problems/Problem 12"
Bluelinfish (talk | contribs) m (→Problem) |
|||
(11 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | In convex quadrilateral <math> ABCD, \angle A \cong \angle C, AB = CD = 180, </math> and <math> AD \neq BC. </math> The perimeter of <math> ABCD </math> is 640. Find <math> \lfloor 1000 \cos A \rfloor. </math> (The notation <math> \lfloor x \rfloor </math> means the greatest integer that is less than or equal to <math> x. </math>) | + | In [[convex]] [[quadrilateral]] <math> ABCD, \angle A \cong \angle C, AB = CD = 180, </math> and <math> AD \neq BC. </math> The [[perimeter]] of <math> ABCD </math> is <math> 640 </math>. Find <math> \lfloor 1000 \cos A \rfloor. </math> (The notation <math> \lfloor x \rfloor </math> means the greatest [[integer]] that is less than or equal to <math> x. </math>) |
== Solution == | == Solution == | ||
− | {{ | + | ===Solution 1=== |
+ | <center><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></center> | ||
+ | |||
+ | By the [[Law of Cosines]] on <math>\triangle ABD</math> at angle <math>A</math> and on <math>\triangle BCD</math> at angle <math>C</math> (note <math>\angle C = \angle A</math>), | ||
+ | |||
+ | <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</cmath> | ||
+ | <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{280}{360} = \dfrac{7}{9} = 0.777 \ldots</math> | ||
+ | |||
+ | <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 == | ||
− | + | {{AIME box|year=2003|n=I|num-b=11|num-a=13}} | |
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:14, 24 November 2019
Problem
In convex quadrilateral and
The perimeter of
is
. 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.