Difference between revisions of "2023 AIME II Problems/Problem 12"
m (→Solution 3 (simplest)) |
m (Added a solution.) |
||
(17 intermediate revisions by 6 users not shown) | |||
Line 5: | Line 5: | ||
==Solution 1== | ==Solution 1== | ||
− | Because <math>M</math> is the midpoint of <math>BC</math>, following from the | + | Because <math>M</math> is the midpoint of <math>BC</math>, following from the Stewart's theorem, <math>AM = 2 \sqrt{37}</math>. |
Because <math>A</math>, <math>B</math>, <math>C</math>, and <math>P</math> are concyclic, <math>\angle BPA = \angle C</math>, <math>\angle CPA = \angle B</math>. | Because <math>A</math>, <math>B</math>, <math>C</math>, and <math>P</math> are concyclic, <math>\angle BPA = \angle C</math>, <math>\angle CPA = \angle B</math>. | ||
Line 14: | Line 14: | ||
<cmath> | <cmath> | ||
\[ | \[ | ||
− | \frac{BQ}{\sin \angle BPA} = \frac{PQ}{\angle PBQ} | + | \frac{BQ}{\sin \angle BPA} = \frac{PQ}{\sin \angle PBQ} |
\] | \] | ||
</cmath> | </cmath> | ||
Line 28: | Line 28: | ||
<cmath> | <cmath> | ||
\[ | \[ | ||
− | \frac{CQ}{\sin \angle CPA} = \frac{PQ}{\angle PCQ} | + | \frac{CQ}{\sin \angle CPA} = \frac{PQ}{\sin \angle PCQ} |
\] | \] | ||
</cmath> | </cmath> | ||
Line 115: | Line 115: | ||
~sigma | ~sigma | ||
− | ==Solution 3 | + | ==Solution 3== |
− | It is clear that <math>BQCP</math> is a parallelogram. By Stewart's Theorem, <math>AM=\sqrt{148}</math>, POP on M tells <math>PM=\frac{49}{\sqrt{148}}</math> | + | It is clear that <math>BQCP</math> is a parallelogram. By Stewart's Theorem, <math>AM=\sqrt{148}</math>, POP on <math>M</math> tells <math>PM=\frac{49}{\sqrt{148}}</math> |
As <math>QM=PM, AQ=AM-PM=\frac{99}{\sqrt{148}}</math> leads to <math>\boxed{247}</math> | As <math>QM=PM, AQ=AM-PM=\frac{99}{\sqrt{148}}</math> leads to <math>\boxed{247}</math> | ||
− | ~bluesoul | + | ~bluesoul (supplemental note: ~Mathavi) |
+ | |||
+ | <math>\textbf{NOTE: Why BQCP is a parallelogram}</math> | ||
+ | |||
+ | It's not actually immediately clear why this is the case. There are two ways to easily show this: | ||
+ | |||
+ | <math>\textbf{Competition solution:}</math> | ||
+ | |||
+ | Notice that the problem statement tells us that point Q is <math>\textit{unique.}</math> EVERY piece of information in the problem statement is intentional, so we should try to use this to our benefit. None of the other solutions do, which is why they are more complicated than they need be. | ||
+ | |||
+ | Consider point Q' s.t. <math>Q'M = MP</math>. Obviously, <math>\angle Q'CP</math> and <math>\angle Q'BP</math> are equal - we have perfect symmetry along line <math>AP</math>. Moreover, <math>BQ'CP</math> is a parallelogram as its diagonals bisect each other. Since point <math>Q</math> is unique, we know that <math>Q' \textit{is } Q</math>. Thus <math>BQCP</math> is a parallelogram. <math>\blacksquare</math> | ||
+ | <math>\newline</math> | ||
+ | <math>\textbf{Rigorous proof (not recommended for competition scenario):}</math> | ||
+ | Consider any quadrilateral <math>ABCD</math> whose diagonals intersect at <math>O</math> s.t. <math>AO = OC</math> and <math>\angle BAD = \angle BCD</math>. We will prove that <math>ABCD</math> is <math>\textit{either a \textbf{parallelogram} or a \textbf{kite}}</math>. | ||
+ | |||
+ | (Note that in our problem, since <math>AP</math> and <math>BC</math> are not orthogonal, (<math>ABC</math> isn't isosceles) this is enough to show that <math>BQCP</math> is a parallelogram). | ||
+ | <math>\newline</math> | ||
+ | -- By same base/same altitude, <math>[ABO] = [CBO]</math> and <math>[ADO] = [CDO] \implies [ABD] = [ABO] + [ADO] = [CBO] + [CDO] = [CBD]. \newline</math> | ||
+ | |||
+ | Therefore: <math>\frac{1}{2} sin(\angle BAD) \overline{AB} \times \overline{AD} = \frac{1}{2} sin(\angle BCD) \overline{CB} \times \overline{CD}.</math> Since <math>\angle BAD = \angle BCD</math>, this reduces to <math>\overline{AB} \times \overline{AD} = \overline{CB} \times \overline{CD}. (E.1) \newline</math> | ||
+ | |||
+ | Let <math>AB = x</math> and <math>AD = y</math>. Then, by <math>(E.1)</math>, <math>CB = kx</math> and <math>CD = \frac{y}{k}</math> for some real <math>k</math>. Then by LoC on <math>\triangle BAD</math> and <math>\triangle BCD</math>: | ||
+ | <math>x^{2} + y^{2} - 2xy cos(\angle BAD) = \overline{BD} = x^{2}k^{2} + \frac{y^{2}}{k^{2}} - 2xy cos(\angle BCD) \newline \implies x^{2} + y^{2} = x^{2}k^{2} + \frac{y^{2}}{k^{2}} \newline \implies (y^{2} - x^{2}k^{2})(k^{2} - 1) = 0.\newline</math> | ||
+ | |||
+ | -- <math>y^{2} - x^{2}k^{2} = 0 \implies y = kx \implies AD = BC</math> and <math>AB = CD \implies</math> <math>ABCD</math> is a parallelogram. | ||
+ | |||
+ | -- <math>k^{2} - 1 = 0 \implies k = 1</math> (<math>k</math> cannot be <math>-1</math>; no negative sided polygons here!) <math>\implies AB = CB</math> and <math>AD = CD \implies</math> <math>ABCD</math> is a kite. <math>\square</math>. ~Mathavi | ||
==Solution 4 (LOS+ coordbash)== | ==Solution 4 (LOS+ coordbash)== | ||
Line 135: | Line 161: | ||
==Solution 5 (similar to 3)== | ==Solution 5 (similar to 3)== | ||
[[File:2023 AIME II 12.png|270px|right]] | [[File:2023 AIME II 12.png|270px|right]] | ||
− | We use the law of Cosine and get <cmath>AB^2 = AM^2 + BM^2 - 2 AM \cdot BM \cos \angle AMB,</cmath> <cmath>AC^2 = | + | We use the law of Cosine and get <cmath>AB^2 = AM^2 + BM^2 - 2 AM \cdot BM \cos \angle AMB,</cmath> <cmath>AC^2 = AM^2 + CM^2 + 2 AM \cdot CM \cos \angle AMB \implies</cmath> <cmath>AM^2 = \frac {AB^2 + AC^2}{2}- BM^2 = \sqrt{148} \approx 12.</cmath> |
We use the power of point <math>M</math> with respect circumcircle <math>\triangle ABC</math> and get | We use the power of point <math>M</math> with respect circumcircle <math>\triangle ABC</math> and get | ||
<cmath>AM \cdot MP = BM \cdot CM = BM^2 \implies</cmath> | <cmath>AM \cdot MP = BM \cdot CM = BM^2 \implies</cmath> | ||
Line 141: | Line 167: | ||
It is clear that if <math>Q = P,</math> then <math>\angle PBQ = \angle PCQ = 0 \implies</math> | It is clear that if <math>Q = P,</math> then <math>\angle PBQ = \angle PCQ = 0 \implies</math> | ||
− | if <math>Q</math> is | + | if <math>Q</math> is symmetric to <math>P</math> with respect <math>M</math> then <math>\angle PBQ = \angle PCQ.</math> |
There exists a <i><b>unique</b></i> point <math>Q</math> on segment <math>\overline{AM}, PM < AM \implies</math> | There exists a <i><b>unique</b></i> point <math>Q</math> on segment <math>\overline{AM}, PM < AM \implies</math> | ||
Line 147: | Line 173: | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | ==Video Solution 1 by SpreadTheMathLove== | + | ==Video Solution 1 (Fastest and Easiest) == |
+ | https://www.youtube.com/watch?v=qm9Sg1tEJJE | ||
+ | |||
+ | ==Video Solution 2 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=k6hEFEVVzMI | https://www.youtube.com/watch?v=k6hEFEVVzMI | ||
Revision as of 21:45, 19 August 2024
Contents
Problem
In with side lengths
and
let
be the midpoint of
Let
be the point on the circumcircle of
such that
is on
There exists a unique point
on segment
such that
Then
can be written as
where
and
are relatively prime positive integers. Find
Solution 1
Because is the midpoint of
, following from the Stewart's theorem,
.
Because ,
,
, and
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
Solution 3
It is clear that is a parallelogram. By Stewart's Theorem,
, POP on
tells
As leads to
~bluesoul (supplemental note: ~Mathavi)
It's not actually immediately clear why this is the case. There are two ways to easily show this:
Notice that the problem statement tells us that point Q is EVERY piece of information in the problem statement is intentional, so we should try to use this to our benefit. None of the other solutions do, which is why they are more complicated than they need be.
Consider point Q' s.t. . Obviously,
and
are equal - we have perfect symmetry along line
. Moreover,
is a parallelogram as its diagonals bisect each other. Since point
is unique, we know that
. Thus
is a parallelogram.
Consider any quadrilateral
whose diagonals intersect at
s.t.
and
. We will prove that
is
.
(Note that in our problem, since and
are not orthogonal, (
isn't isosceles) this is enough to show that
is a parallelogram).
-- By same base/same altitude,
and
Therefore: Since
, this reduces to
Let and
. Then, by
,
and
for some real
. Then by LoC on
and
:
-- and
is a parallelogram.
-- (
cannot be
; no negative sided polygons here!)
and
is a kite.
. ~Mathavi
Solution 4 (LOS+ coordbash)
First, note that by Law of Sines, and that
. Equating the 2 expressions, you get that
. Now drop the altitude from
to
. As it is commonly known that the dropped altitude forms a
and a
triangle, you get the measures of
and
respectively, which are
and
. However, by the inscribed angle theorem, you get that
and that
, respectively. Therefore, by Law of Sines (as previously stated)
.
Now commence coordbashing. Let be the origin, and
be the point
. As
passes through
, which is
, and
, which is
, it has the equation
, so therefore a point on this line can be written as
. As we have the ratio of the lengths, which prompts us to write the lengths in terms of the distance formula, we can just plug and chug it in to get the ratio
. This can be squared to get
. This can be solved to get a solution of
, and an extraneous solution of
which obviously doesn’t work.
Plugging into the line equation gets you
. The distance between this point and
, which is
is
, or simplified to
~dragoon (minor fixes by rhydon516)
Solution 5 (similar to 3)
We use the law of Cosine and get
We use the power of point
with respect circumcircle
and get
It is clear that if
then
if is symmetric to
with respect
then
There exists a unique point on segment
vladimir.shelomovskii@gmail.com, vvsss
Video Solution 1 (Fastest and Easiest)
https://www.youtube.com/watch?v=qm9Sg1tEJJE
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=k6hEFEVVzMI
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.