2023 AIME I Problems/Problem 5

Problem

Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$

Solution 1 (Ptolemy's Theorem)

Ptolemy's theorem states that for cyclic quadrilateral $WXYZ$ , $WX\cdot YZ + XY\cdot WZ = WY\cdot XZ$ .

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 diameter of the circle. 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 (Circle Properties and Half-Angle Formula)

Drop a height from point $P$ to line $\overline{AC}$ and line $\overline{BC}$ . 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 at the center of the circumcircle, call this intersection point $O$ .

Since $OXPY$ is a rectangle, $OX$ is the distance from $P$ to line $\overline{BD}$ . We know that $\tan{\angle{YOX}} = \frac{PX}{XO} = \frac{28}{45}$ by triangle area and given information. Then, notice that the measure of $\angle{OCP}$ is half of $\angle{XOY}$ .

Using the half-angle formula for tangent,

$\begin{align*} \frac{(2*\tan{\angle{OCP}})}{(1-\tan^2{\angle{OCP}})} = \tan{\angle{YOX}} = \frac{28}{45} \\ 14\tan^2{\angle{OCP}} + 45\tan{\angle{OCP}} - 14 = 0 \end{align*}$

Solving the equation above, we get that $\tan{\angle{OCP}} = -7/2$ or $2/7$ . Since this value must be positive, we pick $\frac{2}{7}$ . Then, $\frac{PA}{PC} = 2/7$ (since $\triangle CAP$ is a right triangle with line $\overline{AC}$ also the diameter of the circumcircle) and $PA * PC = 56$ . Solving we get $PA = 4$ , $PC = 14$ , giving us a diagonal of length $\sqrt{212}$ and area $\boxed{106}$ .

~Danielzh

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

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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

Solution 5 (Double Angle)

Notice that $\angle APC$ and $\angle BPD$ are both $90^\circ$

Solution 6 (Similar Triangles)

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        draw (2.8284, -2.8284) -- (2.8284, 2.8284) -- (-2.8284, 2.8284) -- (-2.8284, -2.8284) -- cycle;
        draw (0, 0) node[anchor=north] {$O$};
        draw (-2.8284, -2.8284) node[anchor=north east] {$D$};
        draw (2.8284, -2.8284) node[anchor=north west] {$C$};
        draw (2.8284, 2.8284) node[anchor=south west] {$B$};
        draw (-2.8284, 2.8284) node[anchor=south east] {$A$};
        draw (-0.531, 3.965)
        node[anchor=south] {$P$};
        draw (-2.8284, -2.8284) -- (2.8284, 2.8284) -- (-0.531, 3.965) -- cycle;
        draw (2.8284, -2.8284) -- (-2.8284, 2.8284) -- (-0.531, 3.965) -- cycle;
        draw[dashed] (-0.531, 3.965) -- (1.717, 1.717);
        draw[dashed] (-0.531, 3.965) -- (-2.248, 2.248);
        draw (-2.248, 2.248) node[anchor=north east] {$X$};
        draw (1.717, 1.717) node[anchor=north west] {$Y$};
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Let the center of the circle be $O$ , and the radius of the circle be $r$ . Since $ABCD$ is a rhombus with diagonals $2r$ and $2r$ , its area is $\dfrac{1}{2}(2r)(2r) = 2r^2$ . Since $AC$ and $BD$ are diameters of the circle, $\triangle APC$ and $\triangle BPD$ are right triangles. Let $X$ and $Y$ be the foot of the altitudes to $AC$ and $BD$ , respectively. We have $[\triangle APC] = \frac{1}{2}(PA)(PC) = \frac{1}{2}(PX)(AC),$ so $PX = \dfrac{(PA)(PC)}{AC} = \dfrac{28}{r}$ . Similarly, $[\triangle BPD] = \frac{1}{2}(PB)(PD) = \frac{1}{2}(PY)(PB),$ so $PY = \dfrac{(PB)(PD)}{BD} = \dfrac{45}{r}$ . Since $\triangle APX \sim \triangle PCX,$ $\frac{AX}{PX} = \frac{PX}{CX}$ $\frac{AO - XO}{PX} = \frac{PX}{OC + XO}.$ But $PXOY$ is a rectangle, so $PY = XO$ , and our similarity becomes $\frac{r - PY}{PX} = \frac{PX}{r + PY}.$ Cross multiplying and rearranging gives us $r^2 = PX^2 + PY^2 = \left(\dfrac{28}{r}\right)^2 + \left(\dfrac{45}{r}\right)^2$ , which rearranges to $r^4 = 2809$ . Therefore $[ABCD] = 2r^2 = \boxed{106}$ .

~Cantalon

2023 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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