Difference between revisions of "2019 AIME II Problems/Problem 15"
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It is known that the sides of the orthic triangle are <math>a\cos A,b\cos B,c\cos C</math>, and its angles are <math>\pi-2A</math>,<math>\pi-2B</math>, and <math>\pi-2C</math>. We thus have the three sides of the orthic triangle now. | It is known that the sides of the orthic triangle are <math>a\cos A,b\cos B,c\cos C</math>, and its angles are <math>\pi-2A</math>,<math>\pi-2B</math>, and <math>\pi-2C</math>. We thus have the three sides of the orthic triangle now. | ||
Letting <math>D</math> be the foot of the altitude from <math>A</math>, we have, in <math>\triangle DPQ</math>, | Letting <math>D</math> be the foot of the altitude from <math>A</math>, we have, in <math>\triangle DPQ</math>, | ||
− | <cmath>\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}=27 | + | <cmath>\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}= \frac{27}{35}, \frac{11}{20}.</cmath> |
<cmath>\Rightarrow \cos B=\cos\biggl(\frac{\pi-P}{2}\biggr)=\sin\frac{P}{2}=\sqrt{4/35},</cmath> | <cmath>\Rightarrow \cos B=\cos\biggl(\frac{\pi-P}{2}\biggr)=\sin\frac{P}{2}=\sqrt{4/35},</cmath> | ||
similarly, we get | similarly, we get |
Revision as of 23:56, 7 December 2021
Problem
In acute triangle points
and
are the feet of the perpendiculars from
to
and from
to
, respectively. Line
intersects the circumcircle of
in two distinct points,
and
. Suppose
,
, and
. The value of
can be written in the form
where
and
are positive integers, and
is not divisible by the square of any prime. Find
.
Diagram
Solution 1
First we have , and
by PoP. Similarly,
and dividing these each by
gives
.
It is known that the sides of the orthic triangle are , and its angles are
,
, and
. We thus have the three sides of the orthic triangle now.
Letting
be the foot of the altitude from
, we have, in
,
similarly, we get
To finish,
The requested sum is .
༺\\ crazyeyemoody9❂7 //༻
Solution 2
Let
Therefore
By power of point, we have
Which are simplified to
Or
(1)
Or
Let
Then,
In triangle , by law of cosine
Pluging (1)
Or
Substitute everything by
The quadratic term is cancelled out after simplified
Which gives
Plug back in,
Then
So the final answer is
By SpecialBeing2017
Solution 3
Let and
By power of point, we have
and
Therefore, substituting in the values:
Notice that quadrilateral is cyclic.
From this fact, we can deduce that and
Therefore is similar to
.
Therefore:
Now using Law of Cosines on we get:
Notice
Substituting and Simplifying:
Now we solve for using regular algebra which actually turns out to be very easy.
We get and from the above relations between the variables we quickly determine
,
and
Therefore
So the answer is
By asr41
Solution 4 (Clean)
This solution is directly based of @CantonMathGuy's solution. We start off with a key claim.
Claim. and
.
Proof. Let and
denote the reflections of the orthocenter over points
and
, respectively. Since
and
, we have that
is a rectangle. Then, since
we obtain
(which directly follows from
being cyclic); hence
, or
. Similarly, we can obtain
.
A direct result of this claim is that . Thus, we can set
and
, then applying Power of a Point on
we get
. Also, we can set
and
and once again applying Power of a Point (but this time to
) we get
. Hence,
and the answer is
. ~rocketsri
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.