Difference between revisions of "2023 AIME II Problems/Problem 12"
(Created page with "==Solution== Because <math>M</math> is the midpoint of <math>BC</math>, following from the Steward's theorem, <math>AM = 2 \sqrt{37}</math>. Because <math>A</math>, <math>B<...") |
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Therefore, the answer is <math>99 + 148 = \boxed{\textbf{(247) }}</math>. | Therefore, the answer is <math>99 + 148 = \boxed{\textbf{(247) }}</math>. | ||
+ | ==Solution 2 (== | ||
+ | Define <math>L_1</math> to be the foot of the altitude from <math>A</math> to <math>BC</math>. Furthermore, define <math>L_2</math> to be the foot of the altitude from <math>Q</math> to <math>BC</math>. From here, one can find <math>AL_1=12</math>, either using the 13-14-15 triangle or by calculating the area of <math>ABC</math> two ways. Then, we find <math>BL_1=5</math> and <math>L_1C = 9</math> using Pythagorean theorem. Let <math>QL_2=x</math>. By AA similarity, <math>\triangle{AL_1M}</math> and <math>\triangle{QL_2M}</math> are similar. By similarity ratios, <cmath>\frac{AL_1}{L_1M}=\frac{QL_2}{L_2M}</cmath> | ||
+ | <cmath>\frac{12}{2}=\frac{x}{L_2M}</cmath> | ||
+ | <cmath>L_2M = \frac{x}{6}</cmath> | ||
+ | Thus, <math>BL_2=BM-L_2M=7-\frac{x}{6}</math>. Similarly, <math>CL_2=7+\frac{x}{6}</math>. Now, we angle chase from our requirement to obtain new information. | ||
+ | <cmath>\angle{PBQ}=\angle{PCQ}</cmath> | ||
+ | <cmath>\angle{QCM}+\angle{PCM}=\angle{QBM}+\angle{PBM}</cmath> | ||
+ | <cmath>\angle{QCL_2}+\angle{PCM}=\angle{QBL_2}+\angle{PBM}</cmath> | ||
+ | <cmath>\angle{PCM}-\angle{PBM}=\angle{QBL_2}-\angle{QCL_2}</cmath> | ||
+ | <cmath>\angle{MAB}-\angle{MAC}=\angle{QBL_2}-\angle{QCL_2}\text{(By inscribed angle theorem)}</cmath> | ||
+ | <cmath>(\angle{MAL_1}+\angle{L_1AB})-(\angle{CAL_1}-\angle{MAL_1})=\angle{QBL_2}-\angle{QCL_2}</cmath> | ||
+ | <cmath>2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1}=\angle{QBL_2}-\angle{QCL_2}</cmath> | ||
+ | Take the tangent of both sides to obtain | ||
+ | <cmath>\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\tan(\angle{QBL_2}-\angle{QCL_2})</cmath> | ||
+ | By the definition of the tangent function on right triangles, we have <math>\tan^{-1}{MAL_1}=\frac{7-5}{12}=\frac{1}{6}</math>, <math>\tan^{-1}{CAL_1}=\frac{9}{12}=\frac{3}{4}</math>, and <math>\tan^{-1}{L_1AB}=\frac{5}{12}</math>. By abusing the tangent angle addition formula, we can find that | ||
+ | <cmath>\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\frac{196}{2397}</cmath> | ||
+ | By substituting <math>\tan{\angle{QBL_2}}=\frac{6x}{42-x}</math>, <math>\tan{\angle{QCL_2}}=\frac{6x}{42+x}</math> and using tangent angle subtraction formula we find that | ||
+ | <cmath>x=\frac{147}{37}</cmath> | ||
+ | Finally, using similarity formulas, we can find | ||
+ | <cmath>\frac{AQ}{AM}=\frac{12-x}{x}</cmath>. Plugging in <math>x=\frac{147}{37}</math> and <math>AM=\sqrt{148}</math>, we find that <cmath>AQ=\frac{99}{\sqrt{148}}</cmath> Thus, our final answer is <math>99+148=\boxed{247}</math>. | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 19:58, 16 February 2023
Solution
Because is the midpoint of
, following from the Steward's theorem,
.
Because ,
,
,
are concyclic,
,
.
Denote .
In , following from the law of sines,
Thus,
In , following from the law of sines,
Thus,
Taking , we get
In , following from the law of sines,
Thus, Equations (2) and (3) imply
Next, we compute and
.
We have
We have
Taking (5) and (6) into (4), we get .
Therefore, the answer is
.
Solution 2 (
Define to be the foot of the altitude from
to
. Furthermore, define
to be the foot of the altitude from
to
. From here, one can find
, either using the 13-14-15 triangle or by calculating the area of
two ways. Then, we find
and
using Pythagorean theorem. Let
. By AA similarity,
and
are similar. By similarity ratios,
Thus,
. Similarly,
. Now, we angle chase from our requirement to obtain new information.
Take the tangent of both sides to obtain
By the definition of the tangent function on right triangles, we have
,
, and
. By abusing the tangent angle addition formula, we can find that
By substituting
,
and using tangent angle subtraction formula we find that
Finally, using similarity formulas, we can find
. Plugging in
and
, we find that
Thus, our final answer is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)