Difference between revisions of "2019 AIME II Problems/Problem 15"
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To finish, <cmath>bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}=560\sqrt{14}.</cmath> | To finish, <cmath>bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}=560\sqrt{14}.</cmath> | ||
The requested sum is <math>\boxed{574}</math>. | The requested sum is <math>\boxed{574}</math>. | ||
+ | - crazyeyemoody907 | ||
− | + | Remark: The proof that <math>a \cos A = PQ</math> can be found here: http://www.irmo.ie/5.Orthic_triangle.pdf | |
==Solution 2== | ==Solution 2== | ||
Line 81: | Line 82: | ||
==Solution 4 (Clean)== | ==Solution 4 (Clean)== | ||
− | |||
This solution is directly based of @CantonMathGuy's solution. | This solution is directly based of @CantonMathGuy's solution. | ||
We start off with a key claim. | We start off with a key claim. | ||
− | |||
<i> Claim. </i> <math>XB \parallel AC</math> and <math>YC \parallel AB</math>. | <i> Claim. </i> <math>XB \parallel AC</math> and <math>YC \parallel AB</math>. | ||
<i> Proof. </i> | <i> Proof. </i> | ||
+ | [[File:AIME-II-2019-15.png|350px|right]] | ||
+ | |||
+ | Let <math>E</math> and <math>F</math> denote the reflections of the orthocenter over points <math>P</math> and <math>Q</math>, respectively. Since <math>EF \parallel XY</math> and <cmath>EF = 2 PQ = XP + PQ + QY = XY,</cmath> we have that <math>E X Y F</math> is a rectangle. Then, since <math>\angle XYF = 90^\circ</math> we obtain <math>\angle XBF = 90^\circ</math> (which directly follows from <math>XBYF</math> being cyclic); hence <math>\angle XBQ = \angle AQB</math>, or <math>XB \parallel AQ \Rightarrow XB \parallel AC</math>. | ||
+ | |||
+ | Similarly, we can obtain <math>YC \parallel AB</math>. <math>\ \blacksquare</math> | ||
+ | |||
+ | A direct result of this claim is that <math>\triangle BPX \sim \triangle APQ \sim \triangle CYQ</math>. | ||
− | + | Thus, we can set <math>AP = 5k</math> and <math>BP = 2k</math>, then applying Power of a Point on <math>P</math> we get <math>10 \cdot 40 = 10k^2 \implies k = 2\sqrt{10} \implies AB = 14 \sqrt{10}</math>. Also, we can set <math>AQ = 5l</math> and <math>CQ = 3l</math> and once again applying Power of a Point (but this time to <math>Q</math>) we get | |
− | <math> | + | <math>\phantom{...................}15 \cdot 35 = 15l^2 \implies l = \sqrt{35} \implies AC = 8 \sqrt{35}</math>. |
− | + | Hence, | |
− | + | <math>\phantom{...................}AB \cdot AC = 112 \sqrt{350} = 112 \cdot 5 \sqrt{14} = 560 \sqrt{14}</math> | |
− | + | and the answer is <math>560 + 14 = \boxed{574}</math>. ~rocketsri | |
− | + | ==Solution 5== | |
+ | [[File:2019AIMEIIP15Solution.png|900px]] | ||
+ | '''mathboy282''' | ||
− | + | ==Video Solution by MOP 2024== | |
+ | https://youtu.be/aYV09qIwTqs | ||
− | + | ~r00tsOfUnity | |
==See Also== | ==See Also== |
Latest revision as of 03:11, 7 January 2024
Contents
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
.
- crazyeyemoody907
Remark: The proof that can be found here: http://www.irmo.ie/5.Orthic_triangle.pdf
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 a Point,
Points
and
lie on the circle,
, with diameter
, and pow
, so
Use Law of Cosines in
to get
; since
, this simplifies as
We get
and thus
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
Solution 5
Video Solution by MOP 2024
~r00tsOfUnity
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.