Difference between revisions of "2024 AIME I Problems/Problem 10"
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==Solution 3== | ==Solution 3== | ||
− | Extend sides <math>\overline{AB}</math> and <math>\overline{AC}</math> to points <math>E</math> and <math>F</math>, respectively, such that <math>B</math> and <math>C</math> are the feet of the altitudes in <math>\triangle AEF</math>. Denote the feet of the altitude from <math>A</math> to <math>\overline{EF}</math> as <math>X</math>, and let <math>H</math> denote the orthocenter of <math>\triangle AEF</math>. Call <math>M</math> the midpoint of segment <math>\overline{EF}</math>. By the Three Tangents Lemma, we have that <math>MB</math> and <math>MC</math> are both tangents to <math>(ABC)</math> <math>\implies</math> <math>M = | + | Extend sides <math>\overline{AB}</math> and <math>\overline{AC}</math> to points <math>E</math> and <math>F</math>, respectively, such that <math>B</math> and <math>C</math> are the feet of the altitudes in <math>\triangle AEF</math>. Denote the feet of the altitude from <math>A</math> to <math>\overline{EF}</math> as <math>X</math>, and let <math>H</math> denote the orthocenter of <math>\triangle AEF</math>. Call <math>M</math> the midpoint of segment <math>\overline{EF}</math>. By the Three Tangents Lemma, we have that <math>MB</math> and <math>MC</math> are both tangents to <math>(ABC)</math> <math>\implies</math> <math>M = D</math>, and since <math>M</math> is the midpoint of <math>\overline{EF}</math>, <math>MF = MB</math>. Additionally, by angle chasing, we get that: |
<cmath>\angle ABC \cong \angle AHC \cong \angle EHX</cmath> | <cmath>\angle ABC \cong \angle AHC \cong \angle EHX</cmath> | ||
Also, | Also, | ||
Line 25: | Line 25: | ||
Thus, we can find that the side lengths of <math>\triangle AEF</math> are <math>\frac{250}{11}, \frac{125}{11}, \frac{225}{11}</math>. Then, by Stewart's theorem, <math>AM = \frac{13 \cdot 25}{22}</math>. By Power of a Point, | Thus, we can find that the side lengths of <math>\triangle AEF</math> are <math>\frac{250}{11}, \frac{125}{11}, \frac{225}{11}</math>. Then, by Stewart's theorem, <math>AM = \frac{13 \cdot 25}{22}</math>. By Power of a Point, | ||
<cmath>\overline{MB} \cdot \overline{MB} = \overline{MA} \cdot \overline{MD}</cmath> | <cmath>\overline{MB} \cdot \overline{MB} = \overline{MA} \cdot \overline{MD}</cmath> | ||
− | <cmath>\frac{225}{22} \cdot \frac{225}{22} = \overline{ | + | <cmath>\frac{225}{22} \cdot \frac{225}{22} = \overline{MP} \cdot \frac{13 \cdot 25}{22} \implies \overline{MP} = \frac{225 \cdot 9}{22 \cdot 13}</cmath> |
Thus, | Thus, | ||
− | <cmath> | + | <cmath>AP = AM - MP = \frac{13 \cdot 25}{22} - \frac{225 \cdot 9}{22 \cdot 13} = \frac{100}{13}</cmath> |
Therefore, the answer is <math>\boxed{113}</math>. | Therefore, the answer is <math>\boxed{113}</math>. | ||
Revision as of 20:55, 3 February 2024
Contents
Problem
Let be a triangle inscribed in circle
. Let the tangents to
at
and
intersect at point
, and let
intersect
at
. Find
, if
,
, and
.
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
Well 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(BPC). \\
\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
Video Solution 1 by OmegaLearn.org
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.