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2023 AIME I Problems/Problem 5

Revision as of 12:36, 8 February 2023 by Danielzh (talk | contribs) (→‎Solution 2)

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 $P$ is between $B$ and $C$. Let $PA = a$, $PB = b$, $PC = C$, $PD = d$, and $AB = s$. We have $a^2 + c^2 = AC^2 = 2s^2$, because $AC$ is a diagonal. Similarly, $b^2 + d^2 = 2s^2$. Therefore, $(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112$. Similarly, $(b+d)^2 = 2s^2 + 180$.

By Ptolemy's Theorem on $PCDA$, $as + cs = ds\sqrt{2}$, and therefore $a + c = d\sqrt{2}$. By Ptolemy's on $PBAD$, $bs + ds = as\sqrt{2}$, and therefore $b + d = a\sqrt{2}$. By squaring both equations, we obtain

\[2d^2 = (a+c)^2 = 2s^2 + 112\] \[2a^2 = (b+d)^2 = 2s^2 + 180.\]

Thus, $a^2 = s^2 + 90$, and $d^2 = s^2 + 56$. Plugging these values into $a^2 + c^2 = b^2 + d^2 = 2s^2$, we obtain $c^2 = s^2 - 90$, and $b^2 = s^2 - 56$. Now, we can solve using $a$ and $c$ (though using $b$ and $d$ yields the same solution for $s$).

\[(\sqrt{s^2 + 90})(\sqrt{s^2 - 90}) = ac = 56\] \[(s^2 + 90)(s^2 - 90) = 56^2\] \[s^4 = 90^2 + 56^2 = 106^2\] \[s^2 = 106.\]

The answer is $\boxed{106}$.

~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,

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