Difference between revisions of "1999 AIME Problems/Problem 14"
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== Solution 4 == | == Solution 4 == | ||
− | By splitting the triangle down its altitude, we get a <math>5,12,13</math> and <math>9,12,15</math> triangle. Then <math>\cos\angle{ABC} = \frac{5}{13}</math> and <math>cos\angle{ACB} = \frac{3}{5}</math>. We do this similarly for <math>\cos\angle CAB</math> using Law of Cosine, this is <math>\frac{33}{65}</math>. Now use Law of Sine on triangle <math>PCA</math>. We find that <math>\frac{\sin{\alpha}}{AP} = \frac{\sin{\angle{CPA}}}{15}</math>. Substituting values in we get <math>\frac{\sin{\alpha}}{AP} = \frac{\frac{56}{65}}{15}</math>. Similar expression for triangle <math>BPA</math>, relating expression for <math>AP</math> on that side gets us 2 equation 2 unknown for <math>\alpha</math> and <math>AP</math>. Solving for the tangent of alpha, we get <math>\frac{168}{295} = \boxed{463}</math>. | + | By splitting the triangle down its altitude, we get a <math>5,12,13</math> and <math>9,12,15</math> triangle. Then <math>\cos\angle{ABC} = \frac{5}{13}</math> and <math>\cos\angle{ACB} = \frac{3}{5}</math>. We do this similarly for <math>\cos\angle CAB</math> using Law of Cosine, this is <math>\frac{33}{65}</math>. Now use Law of Sine on triangle <math>PCA</math>. We find that <math>\frac{\sin{\alpha}}{AP} = \frac{\sin{\angle{CPA}}}{15}</math>. Substituting values in we get <math>\frac{\sin{\alpha}}{AP} = \frac{\frac{56}{65}}{15}</math>. Similar expression for triangle <math>BPA</math>, relating expression for <math>AP</math> on that side gets us 2 equation 2 unknown for <math>\alpha</math> and <math>AP</math>. Solving for the tangent of alpha, we get <math>\frac{168}{295} = \boxed{463}</math>. |
== See also == | == See also == |
Revision as of 21:29, 28 May 2021
Problem
Point is located inside triangle
so that angles
and
are all congruent. The sides of the triangle have lengths
and
and the tangent of angle
is
where
and
are relatively prime positive integers. Find
Contents
[hide]Solution
![[asy] real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP("P",P,NW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE); [/asy]](http://latex.artofproblemsolving.com/0/4/f/04fd050352e1902f94f601dda0ad620a7f5a4364.png)
Solution 1
Drop perpendiculars from to the three sides of
and let them meet
and
at
and
respectively.
![[asy] import olympiad; real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP("P",P,SSW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE); /* constructing D,E,F as foot of perps from P */ pair D=foot(P,A,B),E=foot(P,B,C),F=foot(P,C,A); D(MP("D",D,NE)--P--MP("E",E,SSW),dashed);D(P--MP("F",F),dashed); D(rightanglemark(P,E,C,15));D(rightanglemark(P,F,C,15));D(rightanglemark(P,D,A,15)); [/asy]](http://latex.artofproblemsolving.com/f/b/1/fb1fb1dbb7c63477a671fab2b88af58222cc2d43.png)
Let and
. We have that
We can then use the tool of calculating area in two ways
On the other hand,
We still need
though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot:
Adding
gives
Recall that we found that
. Plugging in
, we get
, giving us
for an answer.
Solution 2
Let ,
,
,
,
, and
.
So by the Law of Cosines, we have:
Adding these equations and rearranging, we have:
Now
, by Heron's formula.
Now the area of a triangle, , where
and
are sides on either side of an angle,
. So,
Adding these equations yields:
Dividing
by
, we have:
Thus,
.
Note: In fact, this problem is unfairly easy to those who happen to have learned about Brocard point. The Brocard Angle is given by
Solution 3
Let Then, using Law of Cosines on the three triangles containing vertex
we have
Add the three equations up and rearrange to obtain
Also, using
we have
Divide the two equations to obtain
Solution 4
By splitting the triangle down its altitude, we get a and
triangle. Then
and
. We do this similarly for
using Law of Cosine, this is
. Now use Law of Sine on triangle
. We find that
. Substituting values in we get
. Similar expression for triangle
, relating expression for
on that side gets us 2 equation 2 unknown for
and
. Solving for the tangent of alpha, we get
.
See also
1999 AIME (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 |
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