Difference between revisions of "Mock AIME 1 2010 Problems/Problem 13"
(diagram) |
(→Solution) |
||
Line 58: | Line 58: | ||
</asy> | </asy> | ||
− | <math>\boxed{372}</math>. | + | Let <math>\measuredangle BED=\theta</math>. Because the problem gives us <math>ED^2+EB^2=3050</math>, we think to use the [[Law of Cosines]] in <math>\triangle BED</math>, which yields <math>BD^2=ED^2+EB^2-2ED\cdot EB\cos\theta</math>. Subtituting the values given by the problem, we get <math>28^2=3050-2ED\cdot EB\cos\theta</math>, which gives <math>ED\cdot EB=\tfrac{3050-784}{2\cos\theta}=\tfrac{1133}{\cos\theta}</math>. |
+ | |||
+ | To find another expression for <math>ED\cdot EB</math>, we think of the formula <math>[\triangle BED]=\tfrac12ED\cdot EB\sin\theta</math>. We know that the area of the triangle is <math>\tfrac12\cdot\tfrac{20\sqrt{159}}7\cdot28=40\sqrt{159}</math>. Substituting this in the previous equation for <math>[\triangle BED]</math>, we get that <math>40\sqrt{159}=\tfrac12ED\cdot EB\sin\theta</math>, so <math>ED\cdot EB=\tfrac{80\sqrt{159}}{\sin\theta}</math>. | ||
+ | |||
+ | Setting these two expressions for <math>ED\cdot EB</math> equal to each other reveals that <math>\tfrac{1133}{\cos\theta}=\tfrac{80\sqrt{159}}{\sin\theta}</math>, so <math>\tan\theta=\tfrac{80\sqrt{159}}{1133}</math> by the identity <math>\tan\theta=\tfrac{\sin\theta}{\cos\theta}.</math> | ||
+ | |||
+ | <math>\angle AEB</math> is [[supplementary]] to <math>\angle BED</math>, and <math>\angle ACB</math> is supplementary to <math>\angle AEB</math>, because <math>AEBC</math> is a [[cyclic quadrilateral]]. Thus, <math>\measuredangle BED=\measuredangle ACB=\theta</math>, so <math>\tan\measuredangle ACB=\tan\theta=\tfrac{80\sqrt{159}}{1133}</math>. Thus, <math>a+b+c=80+159+1133=1372</math>, so our answer is <math>\boxed{372}</math>. | ||
== See Also == | == See Also == | ||
{{Mock AIME box|year=2010|n=1|num-b=12|num-a=14}} | {{Mock AIME box|year=2010|n=1|num-b=12|num-a=14}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 17:54, 11 August 2024
Problem
Suppose is inscribed in circle
.
and
are the feet of the altitude from
to
and
to
, respectively. Let
be the intersection of lines
and
, let
be the point of intersection of
and line
distinct from
, and let
be the foot of the perpendicular from
to
. Given that
,
, and
, and that
can be expressed in the form
, where
and
are relatively prime positive integers and
is an integer not divisible by the square of any prime, find the last three digits of
.
Solution
Let . Because the problem gives us
, we think to use the Law of Cosines in
, which yields
. Subtituting the values given by the problem, we get
, which gives
.
To find another expression for , we think of the formula
. We know that the area of the triangle is
. Substituting this in the previous equation for
, we get that
, so
.
Setting these two expressions for equal to each other reveals that
, so
by the identity
is supplementary to
, and
is supplementary to
, because
is a cyclic quadrilateral. Thus,
, so
. Thus,
, so our answer is
.
See Also
Mock AIME 1 2010 (Problems, Source) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |