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2023 AIME I Problems/Problem 5

Revision as of 12:43, 8 February 2023 by Orange quail 9 (talk | contribs) (Added LaTeX to problem statement; added second solution.)

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,

Solution 3 (Analytic geometry)

Denote by $x$ the half length of each side of the square. We put the square to the coordinate plane, with $A = \left( x, x \right)$, $B = \left( - x , x \right)$, $C = \left( - x , - x \right)$, $D = \left( x , - x \right)$.

The radius of the circumcircle of $ABCD$ is $\sqrt{2} x$. Denote by $\theta$ the argument of point $P$ on the circle. Thus, the coordinates of $P$ are $P = \left( \sqrt{2} x \cos \theta , \sqrt{2} x \sin \theta \right)$.

Thus, the equations $PA \cdot PC = 56$ and $PB \cdot PD = 90$ can be written as \begin{align*} \sqrt{\left( \sqrt{2} x \cos \theta - x \right)^2 + \left( \sqrt{2} x \sin \theta - x \right)^2} \cdot \sqrt{\left( \sqrt{2} x \cos \theta + x \right)^2 + \left( \sqrt{2} x \sin \theta + x \right)^2} & = 56 \\ \sqrt{\left( \sqrt{2} x \cos \theta + x \right)^2 + \left( \sqrt{2} x \sin \theta - x \right)^2} \cdot \sqrt{\left( \sqrt{2} x \cos \theta - x \right)^2 + \left( \sqrt{2} x \sin \theta + x \right)^2} & = 90 \end{align*}

These equations can be reformulated as \begin{align*} x^4 \left( 4 - 2 \sqrt{2} \left( \cos \theta + \sin \theta \right) \right) \left( 4 + 2 \sqrt{2} \left( \cos \theta + \sin \theta \right) \right) & = 56^2 \\ x^4 \left( 4 + 2 \sqrt{2} \left( \cos \theta - \sin \theta \right) \right) \left( 4 - 2 \sqrt{2} \left( \cos \theta - \sin \theta \right) \right) & = 90^2 \end{align*}

These equations can be reformulated as \begin{align*} 2 x^4 \left( 1 - 2 \cos \theta \sin \theta \right) & = 28^2 \hspace{1cm} (1) \\ 2 x^4 \left( 1 + 2 \cos \theta \sin \theta \right) & = 45^2 \hspace{1cm} (2) \end{align*}

Taking $\frac{(1)}{(2)}$, by solving the equation, we get \[ 2 \cos \theta \sin \theta = \frac{45^2 - 28^2}{45^2 + 28^2} . \hspace{1cm} (3) \]

Plugging (3) into (1), we get \begin{align*} {\rm Area} \ ABCD & = \left( 2 x \right)^2 \\ & = 4 \sqrt{\frac{28^2}{2 \left( 1 - 2 \cos \theta \sin \theta \right)}} \\ & = 2 \sqrt{45^2 + 28^2} \\ & = 2 \cdot 53 \\ & = \boxed{\textbf{(106) }} . \end{align*}


Solution 4 (Law of Cosines)

WLOG, let $P$ be on minor arc $\overarc {AB}$. Let $r$ and $O$ be the radius and center of the circumcircle respectively, and let $\theta = \angle AOP$.

By the Pythagorean Theorem, the area of the square is $2r^2$. We can use the Law of Cosines on isosceles triangles $\triangle AOP, \, \triangle COP, \, \triangle BOP, \, \triangle DOP$ to get

\begin{align*} PA^2 &= 2r^2(1 - \cos \theta), \\ PC^2 &= 2r^2(1 - \cos (180 - \theta)) = 2r^2(1 + \cos \theta), \\ PB^2 &= 2r^2(1 - \cos (90 - \theta)) = 2r^2(1 - \sin \theta), \\ PD^2 &= 2r^2(1 - \cos (90 + \theta)) = 2r^2(1 + \sin \theta). \end{align*}

Taking the products of the first two and last two equations, respectively, \[56^2 = (PA \cdot PC)^2 = 4r^4(1 - \cos \theta)(1 + \cos \theta) = 4r^4(1 - \cos^2 \theta) = 4r^4 \sin^2 \theta,\] and \[90^2 = (PB \cdot PD)^2 = 4r^4(1 - \sin \theta)(1 + \sin \theta) = 4r^4(1 - \sin^2 \theta) = 4r^4 \cos^2 \theta.\] Adding these equations, \[56^2 + 90^2 = 4r^4,\] so \[2r^2 = \sqrt{56^2+90^2} = 2\sqrt{28^2+45^2} = 2\sqrt{2809} = 2 \cdot 53 = \boxed{106}.\] ~OrangeQuail9

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