Difference between revisions of "2019 AIME II Problems/Problem 15"
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− | Let <math> | + | Let <math>BC=a</math>, <math>AC=b</math>, and <math>AB=c</math>. Let <math>\cos\angle A=k</math>. Then <math>AP=bk</math> and <math>AQ=ck</math>. |
− | + | By Power of a Point theorem, | |
− | + | <cmath>\begin{align} | |
− | + | AP\cdot BP=XP\cdot YP \quad &\Longrightarrow \quad b^2k^2-bck+400=0\\ | |
− | + | AQ\cdot CQ=YQ\cdot XQ \quad &\Longrightarrow \quad c^2k^2-bck+525=0 | |
− | + | \end{align}</cmath> | |
− | + | Thus <math>bck = (bk)^2+400=(ck)^2+525 = u</math>. Then <math>bk=\sqrt{u-400}</math>, <math>ck=\sqrt{u-525}</math>, and | |
− | + | <cmath>k=\sqrt{\frac{(u-400)(u-525)}{u^2}}</cmath> | |
− | + | Use the Law of Cosines in <math>\triangle APQ</math> to get <math>25^2=b^2k^2+c^2k^2-2bck^3 = 2bck-925-2bck^3</math>, which rearranges to <cmath>775=bck - k^2\cdot bck = u-\frac{(u-400)(u-525)}{u}</cmath>Upon simplification, this reduces to a linear equation in <math>u</math>, with solution <math>u=1400</math>. Then <cmath>AB\cdot AC = bc = \frac 1{k}\cdot bck = \frac{u^2}{\sqrt{(u-400)(u-525)}}=560 \sqrt{14}</cmath> | |
− | + | So the final answer is <math>560 + 14 = \boxed{574}</math> | |
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− | So the final answer is <math>560 + 14 = \boxed{574} </math> | ||
By SpecialBeing2017 | By SpecialBeing2017 |
Revision as of 12:01, 17 January 2022
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 ,
, and
. Let
. Then
and
.
By Power of a Point theorem,
Thus
. Then
,
, and
Use the Law of Cosines in
to get
, which rearranges to
Upon simplification, this reduces to a linear equation in
, with solution
. 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.