Difference between revisions of "2023 AIME II Problems/Problem 12"
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Taking (5) and (6) into (4), we get <math>AQ = \frac{99}{\sqrt{148}}</math>. | Taking (5) and (6) into (4), we get <math>AQ = \frac{99}{\sqrt{148}}</math>. | ||
Therefore, the answer is <math>99 + 148 = \boxed{\textbf{(247) }}</math>. | Therefore, the answer is <math>99 + 148 = \boxed{\textbf{(247) }}</math>. | ||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution 2 | + | ==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> | 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>\frac{12}{2}=\frac{x}{L_2M}</cmath> | ||
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Finally, using similarity formulas, we can find | 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>. | <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>. | ||
− | ~ | + | ~sigma |
Revision as of 19:59, 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
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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
.
~sigma