Difference between revisions of "2024 AIME I Problems/Problem 10"
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==Problem== | ==Problem== | ||
− | Let <math>ABC</math> be a triangle inscribed in circle <math>\omega</math>. Let the tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at point <math>D</math>, and let <math>\overline{AD}</math> intersect <math>\omega</math> at <math>P</math>. | + | Let <math>ABC</math> be a triangle inscribed in circle <math>\omega</math>. Let the tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at point <math>D</math>, and let <math>\overline{AD}</math> intersect <math>\omega</math> at <math>P</math>. If <math>AB=5</math>, <math>BC=9</math>, and <math>AC=10</math>, <math>AP</math> can be written as the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime integers. Find <math>m + n</math>. |
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | import olympiad; | ||
+ | |||
+ | unitsize(15); | ||
+ | |||
+ | pair A, B, C, D, E, F, P, O; | ||
+ | |||
+ | C = origin; A = (10,0); B = (7.8, 4.4899); | ||
+ | draw(A--B--C--cycle); draw(A..B..C..cycle, red+dotted); | ||
+ | |||
+ | O = circumcenter(A, B, C); | ||
+ | |||
+ | E = rotate(90,B) * (O); | ||
+ | F = rotate(90,C) * (O); | ||
+ | |||
+ | D = IP(B..E + (B-E)*4, C..F + (C-F)*-3); | ||
+ | |||
+ | draw(B--D--C--D--A); | ||
+ | |||
+ | P = IP(D..A, A..B..C); | ||
+ | |||
+ | dot(A); dot(B); dot(C); dot(D); dot(P); | ||
+ | label("$A$", A, dir(335)); | ||
+ | label("$B$", B, dir(65)); | ||
+ | label("$C$", C, dir(200)); | ||
+ | label("$D$", D, dir(135)); | ||
+ | label("$P$", P, dir(235)); | ||
+ | </asy> | ||
==Solution 1== | ==Solution 1== | ||
Line 9: | Line 39: | ||
==Solution 2== | ==Solution 2== | ||
− | + | We know <math>AP</math> is the symmedian, which implies <math>\triangle{ABP}\sim \triangle{AMC}</math> where <math>M</math> is the midpoint of <math>BC</math>. By Appolonius theorem, <math>AM=\frac{13}{2}</math>. Thus, we have <math>\frac{AP}{AC}=\frac{AB}{AM}, AP=\frac{100}{13}\implies \boxed{113}</math> | |
~Bluesoul | ~Bluesoul |
Latest revision as of 19:47, 18 August 2024
Contents
Problem
Let be a triangle inscribed in circle
. Let the tangents to
at
and
intersect at point
, and let
intersect
at
. If
,
, and
,
can be written as the form
, where
and
are relatively prime integers. Find
.
Diagram
Solution 1
From the tangency condition we have . With LoC we have
and
. Then,
. Using LoC we can find
:
. Thus,
. By Power of a Point,
so
which gives
. Finally, we have
.
~angie.
Solution 2
We know is the symmedian, which implies
where
is the midpoint of
. By Appolonius theorem,
. Thus, we have
~Bluesoul
Solution 3
Extend sides and
to points
and
, respectively, such that
and
are the feet of the altitudes in
. Denote the feet of the altitude from
to
as
, and let
denote the orthocenter of
. Call
the midpoint of segment
. By the Three Tangents Lemma, we have that
and
are both tangents to
, and since
is the midpoint of
,
. Additionally, by angle chasing, we get that:
Also,
Furthermore,
From this, we see that
with a scale factor of
. By the Law of Cosines,
Thus, we can find that the side lengths of
are
. Then, by Stewart's theorem,
. By Power of a Point,
Thus,
Therefore, the answer is
.
~mathwiz_1207
Solution 4 (LoC spam)
Connect lines and
. From the angle by tanget formula, we have
. Therefore by AA similarity,
. Let
. Using ratios, we have
Similarly, using angle by tangent, we have
, and by AA similarity,
. By ratios, we have
However, because
, we have
so
Now using Law of Cosines on
in triangle
, we have
Solving, we find
. Now we can solve for
. Using Law of Cosines on
we have
\begin{align*}
81&=x^2+4x^2-4x^2\cos(180-\angle BAC) \\
&= 5x^2+4x^2\cos(BAC). \\
\end{align*}
Solving, we get Now we have a system of equations using Law of Cosines on
and
,
Solving, we find , so our desired answer is
.
~evanhliu2009
Solution 5
Following from the law of cosines, we can easily get ,
,
.
Hence, ,
,
.
Thus,
.
Denote by the circumradius of
.
In
, following from the law of sines, we have
.
Because and
are tangents to the circumcircle
,
and
.
Thus,
.
In , we have
and
.
Thus, following from the law of cosines, we have
\begin{align*} AD & = \sqrt{OA^2 + OD^2 - 2 OA \cdot OD \cos \angle AOD} \\ & = \frac{26 \sqrt{14}}{33} R. \end{align*}
Following from the law of cosines,
\begin{align*} \cos \angle OAD & = \frac{AD^2 + OA^2 - OD^2}{2 AD \cdot OA} \\ & = \frac{8 \sqrt{14}}{39} . \end{align*}
Therefore,
\begin{align*} AP & = 2 OA \cos \angle OAD \\ & = \frac{100}{13} . \end{align*}
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 1 by OmegaLearn.org
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.