Difference between revisions of "2024 AIME I Problems/Problem 10"
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− | From the tangency condition we have <math>\let\angle | + | From the tangency condition we have <math>\let\angle BCD = \let\angle CBD = \let\angle A</math>. With LoC we have <math>\cos(A) = \frac{25+100-81}{2*5*10} = \frac{11}{25}</math> and <math>\cos(B) = \frac{81+25-100}{2*9*5} = \frac{1}{15}</math>. Then, <math>CD = \frac{\frac{9}{2}}{\cos(A)} = \frac{225}{22}</math>. Using LoC we can find <math>AD</math>: <math>AD^2 = AC^2 + CD^2 - 2(AC)(CD)\cos(A+C) = 10^2+(\frac{225}{22})^2 + 2(10)\frac{225}{22}\cos(B) = 100 + \frac{225^2}{22^2} + 2(10)\frac{225}{22}*\frac{1}{15} = \frac{5^4*13^2}{484}</math>. Thus, <math>AD = \frac{5^2*13}{22}</math>. By Power of a Point, <math>DP*AD = CD^2</math> so <math>DP*\frac{5^2*13}{22} = (\frac{225}{22})^2</math> which gives <math>DP = \frac{5^2*9^2}{13*22}</math>. Finally, we have <math>AP = AD - DP = \frac{5^2*13}{22} - \frac{5^2*9^2}{13*22} = \frac{100}{13} \rightarrow \boxed{113}</math>. |
~angie. | ~angie. |
Revision as of 21:17, 6 February 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
.
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 |
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