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> | + | 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>. |
==Solution 1== | ==Solution 1== | ||
Line 9: | Line 9: | ||
==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 | ||
+ | |||
+ | ==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 = 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> | ||
+ | Also, | ||
+ | <cmath>\angle EHX = 90 ^\circ - \angle HEF = 90 ^\circ - (90 ^\circ - \angle AFE) = \angle AFE</cmath> | ||
+ | Furthermore, | ||
+ | <cmath>AB = AF \cdot \cos(A)</cmath> | ||
+ | From this, we see that <math>\triangle ABC \sim \triangle AFE</math> with a scale factor of <math>\cos(A)</math>. By the Law of Cosines, | ||
+ | <cmath>\cos(A) = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}</cmath> | ||
+ | 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{MP}</cmath> | ||
+ | <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, | ||
+ | <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>. | ||
+ | |||
+ | ~mathwiz_1207 | ||
+ | |||
+ | ==Solution 4 (LoC spam)== | ||
+ | Connect lines <math>\overline{PB}</math> and <math>\overline{PC}</math>. From the angle by tanget formula, we have <math>\angle PBD = \angle DAB</math>. Therefore by AA similarity, <math>\triangle PBD \sim \triangle BAD</math>. Let <math>\overline{BP} = x</math>. Using ratios, we have <cmath>\frac{x}{5}=\frac{BD}{AD}.</cmath> Similarly, using angle by tangent, we have <math>\angle PCD = \angle DAC</math>, and by AA similarity, <math>\triangle CPD \sim \triangle ACD</math>. By ratios, we have <cmath>\frac{PC}{10}=\frac{CD}{AD}.</cmath> However, because <math>\overline{BD}=\overline{CD}</math>, we have <cmath>\frac{x}{5}=\frac{PC}{10},</cmath> so <math>\overline{PC}=2x.</math> Now using Law of Cosines on <math>\angle BAC</math> in triangle <math>\triangle ABC</math>, we have <cmath>9^2=5^2+10^2-100\cos(\angle BAC).</cmath> Solving, we find <math>\cos(\angle BAC)=\frac{11}{25}</math>. Now we can solve for <math>x</math>. Using Law of Cosines on <math>\triangle BPC,</math> 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 <math>x=\frac{45}{13}.</math> Now we have a system of equations using Law of Cosines on <math>\triangle BPA</math> and <math>\triangle CPA</math>, <cmath>AP^2=5^2+\left(\frac{45}{13}\right)^2 -(10) \left(\frac{45}{13} \right)\cos(ABP)</cmath> | ||
+ | <cmath>AP^2=10^2+4 \left(\frac{45}{13} \right)^2 + (40) \left(\frac{45}{13} \right)\cos(ABP).</cmath> | ||
+ | |||
+ | Solving, we find <math>\overline{AP}=\frac{100}{13}</math>, so our desired answer is <math>100+13=\boxed{113}</math>. | ||
+ | |||
+ | ~evanhliu2009 | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Following from the law of cosines, we can easily get <math>\cos A = \frac{11}{25}</math>, <math>\cos B = \frac{1}{15}</math>, <math>\cos C = \frac{13}{15}</math>. | ||
+ | |||
+ | Hence, <math>\sin A = \frac{6 \sqrt{14}}{25}</math>, <math>\cos 2C = \frac{113}{225}</math>, <math>\sin 2C = \frac{52 \sqrt{14}}{225}</math>. | ||
+ | Thus, <math>\cos \left( A + 2C \right) = - \frac{5}{9}</math>. | ||
+ | |||
+ | Denote by <math>R</math> the circumradius of <math>\triangle ABC</math>. | ||
+ | In <math>\triangle ABC</math>, following from the law of sines, we have <math>R = \frac{BC}{2 \sin A} = \frac{75}{4 \sqrt{14}}</math>. | ||
+ | |||
+ | Because <math>BD</math> and <math>CD</math> are tangents to the circumcircle <math>ABC</math>, <math>\triangle OBD \cong \triangle OCD</math> and <math>\angle OBD = 90^\circ</math>. | ||
+ | Thus, <math>OD = \frac{OB}{\cos \angle BOD} = \frac{R}{\cos A}</math>. | ||
+ | |||
+ | In <math>\triangle AOD</math>, we have <math>OA = R</math> and <math>\angle AOD = \angle BOD + \angle AOB = A + 2C</math>. | ||
+ | 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 <math>100 + 13 = \boxed{\textbf{(113) }}</math>. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution 1 by OmegaLearn.org== | ||
+ | https://youtu.be/heryP002bp8 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/RawwQmVYyaw | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.