Difference between revisions of "2023 AIME I Problems/Problem 5"
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− | ==Solution 2== | + | ==Solution 2 (Trigonometry)== |
Drop a height from point P to line AC and BC. Call these two points to be X and Y, respectively. Notice that the intersection of the diagonals of square ABCD meets at a right angle and at the center of the circumcircle, call this intersection point O. | Drop a height from point P to line AC and BC. Call these two points to be X and Y, respectively. Notice that the intersection of the diagonals of square ABCD meets at a right angle and at the center of the circumcircle, call this intersection point O. | ||
− | + | Since OXPY is a rectangle, OX is the distance from P to line BD. We know the that tan(YOX) = PX/XO = 28/45 by triangle area and given information. Then, notice that the measure of angle OCP is half of the angle of using half angle formula, |
Revision as of 13:36, 8 February 2023
Problem (not official; when the official problem statement comes out, please update this page; to ensure credibility until the official problem statement comes out, please add an O if you believe this is correct and add an X if you believe this is incorrect):
Let there be a circle circumscribing a square ABCD, and let P be a point on the circle. PA*PC = 56, PB*PD = 90. What is the area of the square?
Solution
We may assume that is between
and
. Let
,
,
,
, and
. We have
, because
is a diagonal. Similarly,
. Therefore,
. Similarly,
.
By Ptolemy's Theorem on ,
, and therefore
. By Ptolemy's on
,
, and therefore
. By squaring both equations, we obtain
Thus, , and
. Plugging these values into
, we obtain
, and
. Now, we can solve using
and
(though using
and
yields the same solution for
).
The answer is .
~mathboy100
Solution 2 (Trigonometry)
Drop a height from point P to line AC and BC. Call these two points to be X and Y, respectively. Notice that the intersection of the diagonals of square ABCD meets at a right angle and at the center of the circumcircle, call this intersection point O. Since OXPY is a rectangle, OX is the distance from P to line BD. We know the that tan(YOX) = PX/XO = 28/45 by triangle area and given information. Then, notice that the measure of angle OCP is half of the angle of using half angle formula,