Difference between revisions of "2023 AIME I Problems/Problem 12"
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Draw line segments from P to points A, B, and C. And label the angle measure of <math>\angle{BFP}</math>, <math>\angle{CDP}</math>, and <math>\angle{AEP}</math> to be <math>\alpha</math> | Draw line segments from P to points A, B, and C. And label the angle measure of <math>\angle{BFP}</math>, <math>\angle{CDP}</math>, and <math>\angle{AEP}</math> to be <math>\alpha</math> | ||
− | Using Law of Cosines (note that <math>cos{\angle{AFP}}=cos{\angle{BDP}}=cos{\angle{CEP}}=cos{\ | + | Using Law of Cosines (note that <math>cos{\angle{AFP}}=cos{\angle{BDP}}=cos{\angle{CEP}}=cos{180^\circ-\alpha}=-cos{\alpha}</math>) |
<cmath> | <cmath> | ||
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Therefore, we want to solve for <math>DP+EP+FP</math> | Therefore, we want to solve for <math>DP+EP+FP</math> | ||
− | Notice that <math>\angle{APC}=\angle{APC}=\angle{APC}=\ | + | Notice that <math>\angle{APC}=\angle{APC}=\angle{APC}=120^\circ</math> |
We can use Law of Cosines again to solve for the sides of DEF, which have side lengths of 13, 42, and 35, and area <math>120\sqrt{3}</math>. | We can use Law of Cosines again to solve for the sides of DEF, which have side lengths of 13, 42, and 35, and area <math>120\sqrt{3}</math>. |
Revision as of 01:06, 9 February 2023
Problem 12
Let be an equilateral triangle with side length
. Points
,
, and
lie on sides
,
, and
, respectively, such that
,
, and
. A unique point
inside
has the property that
Find
.
Solution
Denote .
In , we have
.
Thus,
Taking the real and imaginary parts, we get
In , analogous to the analysis of
above, we get
Taking , we get
Taking , we get
Taking , we get
Therefore,
Therefore, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (no trig)
Drop the perpendiculars from to
,
,
, and call them
and
respectively. This gives us three similar right triangles
,
, and
The sum of the perpendiculars to a point within an equilateral triangle is always constant, so we have that
The sum of the lengths of the alternating segments split by the perpendiculars from a point within an equilateral triangle is always equal to half the perimeter, so
which means that
Finally,
Thus,
~anon
Solution 3 (LOC)
This solution is heavily inspired by AIME 1999 Problem 14 Solution 2 (a similar question)
Draw line segments from P to points A, B, and C. And label the angle measure of ,
, and
to be
Using Law of Cosines (note that )
We can perform this operation: (1) - (2) + (3) - (4) + (5) - (6)
Leaving us with (after combining and simplifying)
Therefore, we want to solve for
Notice that
We can use Law of Cosines again to solve for the sides of DEF, which have side lengths of 13, 42, and 35, and area .
Label the lengths of ,
, and
to be x, y, and z.
Therefore, using the area formula,
In addition, we know that
By using Law of Cosines for ,
, and
respectively
Because we want , which is
, we see that
So plugging the results back into the equation before, we get
Giving us
~Danielzh
See also
2023 AIME I (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.