Difference between revisions of "2005 AIME II Problems/Problem 14"
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Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. 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 <math>\boxed{463}</math>. | Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. 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 <math>\boxed{463}</math>. | ||
+ | |||
+ | ==Solution 3 (LoC and LoS bash)== | ||
+ | Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have | ||
+ | <cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{BEA}}</cmath> | ||
+ | As a result, our goal is to find <math>\sin{BEA}</math> and <math>\sin{\theta}</math> (we already know <math>AB</math>). | ||
+ | |||
+ | Let the foot of the altitude from <math>A</math> to <math>BC</math> be <math>H</math>. By law of cosines on <math>\triangle ABC</math> we have | ||
+ | <cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath> | ||
+ | It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}</math>. | ||
+ | |||
+ | Note that by PT on <math>\triangle AHD</math> we have that <math>AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}</math>. By Law of Sines on <math>\triangle ADC</math> (where we square everything to avoid taking the square root) we see | ||
+ | <cmath>\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}</cmath> | ||
+ | |||
+ | |||
== See also == | == See also == |
Revision as of 00:13, 28 July 2017
Contents
[hide]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
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.