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

(→‎Solution 2)
Line 26: Line 26:
 
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,
 
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 <math>x</math> the half length of each side of the square.
 +
We put the square to the coordinate plane, with <math>A = \left( x, x \right)</math>, <math>B = \left( - x , x \right)</math>, <math>C = \left( - x , - x \right)</math>, <math>D = \left( x , - x \right)</math>.
 +
 +
The radius of the circumcircle of <math>ABCD</math> is <math>\sqrt{2} x</math>.
 +
Denote by <math>\theta</math> the argument of point <math>P</math> on the circle.
 +
Thus, the coordinates of <math>P</math> are <math>P = \left( \sqrt{2} x \cos \theta , \sqrt{2} x \sin \theta \right)</math>.
 +
 +
Thus, the equations <math>PA \cdot PC = 56</math> and <math>PB \cdot PD = 90</math> can be written as
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
These equations can be reformulated as
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
These equations can be reformulated as
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Taking <math>\frac{(1)}{(2)}</math>, by solving the equation, we get
 +
<cmath>
 +
\[
 +
2 \cos \theta \sin \theta = \frac{45^2 - 28^2}{45^2 + 28^2} . \hspace{1cm} (3)
 +
\]
 +
</cmath>
 +
 +
Plugging (3) into (1), we get
 +
<cmath>
 +
\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*}
 +
</cmath>

Revision as of 13:39, 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

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*}

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