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Difference between revisions of "2023 AIME I Problems/Problem 5"

(Created page with "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...")
 
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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?
 
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?
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==Solution==
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We may assume that <math>P</math> is between <math>B</math> and <math>C</math>. Let <math>PA = a</math>, <math>PB = b</math>, <math>PC = C</math>, <math>PD = d</math>, and <math>AB = s</math>. We have <math>a^2 + c^2 = AC^2 = 2s^2</math>, because <math>AC</math> is a diagonal. Similarly, <math>b^2 + d^2 = 2s^2</math>. Therefore, <math>(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112</math>. Similarly, <math>(b+d)^2 = 2s^2 + 180</math>.
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By Ptolemy's Theorem on <math>PCDA</math>, <math>as + cs = ds\sqrt{2}</math>, and therefore <math>a + c = d\sqrt{2}</math>. By Ptolemy's on <math>PBAD</math>, <math>bs + ds = as\sqrt{2}</math>, and therefore <math>b + d = a\sqrt{2}</math>. By squaring both equations, we obtain
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<cmath>2d^2 = (a+c)^2 = 2s^2 + 112</cmath>
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<cmath>2a^2 = (b+d)^2 = 2s^2 + 180.</cmath>
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Thus, <math>a^2 = s^2 + 90</math>, and <math>d^2 = s^2 + 56</math>. Plugging these values into <math>a^2 + c^2 = b^2 + d^2 = 2s^2</math>, we obtain <math>c^2 = s^2 - 90</math>, and <math>b^2 = s^2 - 56</math>. Now, we can solve using <math>a</math> and <math>c</math> (though using <math>b</math> and <math>d</math> yields the same solution for <math>s</math>).
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<cmath>(\sqrt{s^2 + 90})(\sqrt{s^2 - 90}) = ac = 56</cmath>
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<cmath>(s^2 + 90)(s^2 - 90) = 56^2</cmath>
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<cmath>s^4 = 90^2 + 56^2 = 106^2</cmath>
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<cmath>s^2 = 106.</cmath>
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The answer is <math>\boxed{106}</math>.
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~mathboy100

Revision as of 11:24, 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 $P$ 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

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