Difference between revisions of "2005 AIME II Problems/Problem 14"
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We have that <math>\sin{180^{\circ} - (\theta + \angle B)} = \sin{\angle BEA} = \sin{\theta + \angle B}</math>. Success! We know <math>\sin{\theta}</math> and <math>\sin{\angle B}</math> already. Applying the <math>\sin</math> addition formula we see | We have that <math>\sin{180^{\circ} - (\theta + \angle B)} = \sin{\angle BEA} = \sin{\theta + \angle B}</math>. Success! We know <math>\sin{\theta}</math> and <math>\sin{\angle B}</math> already. Applying the <math>\sin</math> addition formula we see | ||
− | <cmath>\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}</cmath> | + | <cmath>\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}</cmath> |
+ | This is the last stretch! Applying Law of Sines a final time on <math>\triangle BEA</math> we see | ||
+ | <cmath>\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}</cmath> | ||
== See also == | == See also == |
Revision as of 01:29, 28 July 2017
Contents
Problem
In triangle and
Point
is on
with
Point
is on
such that
Given that
where
and
are relatively prime positive integers, find
Solution 1
![[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]](http://latex.artofproblemsolving.com/f/5/3/f53f3016596f4d0c3c6dc23b6e0b5b41fabe85f2.png)
By the Law of Sines and since , we have
Substituting our knowns, we have . The answer is
.
Solution 2 (Similar Triangles)
Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively.
From here, we can use Heron's Formula to find the altitude. The area of the triangle is = 84. We can then use similar triangles with triangle AQC and triangle DSC to find DS=
. Consequently, from Pythagorean theorem, SC =
and AS = 14-SC =
. We can also use pythagorean triangle on triangle AQB to determine that BQ =
.
Label AR as y and RE as x. RB then equals 13-y. Then, we have two similar triangles.
Firstly: . From there, we have
.
Next: . From there, we have
.
Solve the system to get and
. Notice that 463 is prime, so even though we use pythagorean theorem on x and 13-y, the denominator won't change. The answer we desire is
.
Solution 3 (LoC and LoS bash)
Let . Note by Law of Sines on
we have
As a result, our goal is to find
and
(we already know
).
Let the foot of the altitude from to
be
. By law of cosines on
we have
It follows that
and
.
Note that by PT on we have that
. By Law of Sines on
(where we square everything to avoid taking the square root) we see
How are we going to find
though?
and
are in the same triangle. Applying Law of Sines on
we see that
,
, and
are all in the same triangle. We know they add up to
. There's a good chance we can exploit this using the identity
.
We have that . Success! We know
and
already. Applying the
addition formula we see
This is the last stretch! Applying Law of Sines a final time on
we see
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.