Difference between revisions of "1999 AIME Problems/Problem 14"
(→Solution: Lengths of 14 and 15 were switched, attempted to fix asy code) |
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D(A--MP("P",P,NW)--B);D(P--C); | 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)); | 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(" | + | MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE); |
</asy></center> | </asy></center> | ||
<!--This image does exist now: [[Image:1999_AIME-14.png]]--> | <!--This image does exist now: [[Image:1999_AIME-14.png]]--> | ||
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D(A--MP("P",P,SSW)--B);D(P--C); | 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)); | 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(" | + | 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 */ | /* constructing D,E,F as foot of perps from P */ |
Revision as of 23:15, 17 December 2013
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
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,
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.