2023 AIME I Problems/Problem 5

1 Problem
2 Solution 1 (Ptolemy's Theorem)
3 Solution 2 (Heights and Half-Angle Formula)
4 Solution 3 (Analytic Geometry)
5 Solution 4 (Law of Cosines)
6 Solution 5 (Similar Triangles)
7 Solution 6 (Subtended Chords)
8 Solution 7 (Areas and Pythagorean Theorem)
9 Solution 8 (Coordinates and algebraic manipulation)
10 See also

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 $\begin{alignat*}{8} 2d^2 &= (a+c)^2 &&= 2s^2 + 112, \\ 2a^2 &= (b+d)^2 &&= 2s^2 + 180. \end{alignat*}$ 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 $s$ ). $\begin{align*} ac = (\sqrt{s^2 + 90})(\sqrt{s^2 - 90}) &= 56 \\ (s^2 + 90)(s^2 - 90) &= 56^2 \\ s^4 = 90^2 + 56^2 &= 106^2 \\ s^2 = \boxed{106}. \end{align*}$ ~mathboy100

Solution 2 (Heights 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 $\overline{AC}$ 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 (Similar Triangles)

Someone please help render this

\begin{center}

   \begin{tikzpicture}
       \draw (0,0) circle (4cm);
       \draw (2.8284, -2.8284) -- (2.8284, 2.8284) -- (-2.8284, 2.8284) -- (-2.8284, -2.8284) -- cycle;
       \draw (0, 0) node[anchor=north] {};
       \draw (-2.8284, -2.8284) node[anchor=north east] {};
       \draw (2.8284, -2.8284) node[anchor=north west] {};
       \draw (2.8284, 2.8284) node[anchor=south west] {};
       \draw (-2.8284, 2.8284) node[anchor=south east] {};
       \draw (-0.531, 3.965)
       node[anchor=south] {};
       \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] {};
       \draw (1.717, 1.717) node[anchor=north west] {};
   \end{tikzpicture}

\end{center}

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

Solution 6 (Subtended Chords)

First draw a diagram. $[asy] pair A, B, C, D, O, P; A = (0,sqrt(106)); B = (0,0); C = (sqrt(106),0); D = (sqrt(106),sqrt(106)); O = (sqrt(106)/2, sqrt(106)/2); P = intersectionpoint(circle(A, sqrt(212)*sin(atan(28/45)/2)), circle(O, sqrt(212)/2)); draw(A--B--C--D--cycle); draw(circle(O, sqrt(212)/2)); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, NE); label("$P$", P, NW); label("$O$", O, SW); label("$\theta$", O, dir(120)*5); dot(P); dot(O); draw(P--A--C--cycle, red); draw(P--B--D--cycle, blue); draw(P--O); draw(anglemark(P,O,A,30)); [/asy]$ Let's say that the radius is $r$ . Then the area of the $ABCD$ is $(\sqrt2r)^2 = 2r^2$ Using the formula for the length of a chord subtended by an angle, we get $PA = 2r\sin\left(\dfrac{\theta}2\right)$ $PC = 2r\sin\left(\dfrac{180-\theta}2\right) = 2r\sin\left(90 - \dfrac{\theta}2\right) = 2r\cos\left(\dfrac{\theta}2\right)$ Multiplying and simplifying these 2 equations gives $PA \cdot PC = 4r^2 \sin \left(\dfrac{\theta}2 \right) \cos \left(\dfrac{\theta}2 \right) = 2r^2 \sin\left(\theta \right) = 56$ Similarly $PB = 2r\sin\left(\dfrac{90 +\theta}2\right)$ and $PD =2r\sin\left(\dfrac{90 -\theta}2\right)$ . Again, multiplying gives $PB \cdot PD = 4r^2 \sin\left(\dfrac{90 +\theta}2\right) \sin\left(\dfrac{90 -\theta}2\right) = 4r^2 \sin\left(90 -\dfrac{90 -\theta}2\right) \sin\left(\dfrac{90 -\theta}2\right)$ $=4r^2 \sin\left(\dfrac{90 -\theta}2\right) \cos\left(\dfrac{90 -\theta}2\right) = 2r^2 \sin\left(90 - \theta \right) = 2r^2 \cos\left(\theta \right) = 90$ Dividing $2r^2 \sin \left(\theta \right)$ by $2r^2 \cos \left( \theta \right)$ gives $\tan \left(\theta \right) = \dfrac{28}{45}$ , so $\theta = \tan^{-1} \left(\dfrac{28}{45} \right)$ . Pluging this back into one of the equations, gives $2r^2 = \dfrac{90}{\cos\left(\tan^{-1}\left(\dfrac{28}{45}\right)\right)}$ If we imagine a 28-45-53 right triangle, we see that if 28 is opposite and 45 is adjacent, $\cos\left(\theta\right) = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{45}{53}$ . Now we see that $2r^2 = \dfrac{90}{\frac{45}{53}} = \boxed{106}$ ~Voldemort101

Solution 7 (Areas and Pythagorean Theorem)

By the Inscribed Angle Theorem, we conclude that $\triangle PAC$ and $\triangle PBD$ are right triangles.

Let the brackets denote areas. We are given that $\begin{alignat*}{8} 2[PAC] &= PA \cdot PC &&= 56, \\ 2[PBD] &= PB \cdot PD &&= 90. \end{alignat*}$ Let $O$ be the center of the circle, $X$ be the foot of the perpendicular from $P$ to $\overline{AC},$ and $Y$ be the foot of the perpendicular from $P$ to $\overline{BD},$ as shown below.

(Diagram available soon.)

Let $d$ be the diameter of $\odot O.$ It follows that $\begin{alignat*}{8} 2[PAC] &= d\cdot PX &&= 56, \\ 2[PBD] &= d\cdot PY &&= 90. \end{alignat*}$ Moreover, note that $OXPY$ is a rectangle. The Pythagorean Theorem gives $PX^2+PY^2=(PO)^2.$ We rewrite this equation in terms of $d:$ $\left(\frac{56}{d}\right)^2+\left(\frac{90}{d}\right)^2=\left(\frac d2\right)^2,$ from which $d^2=212.$

Finally, we have $[ABCD] = \frac{d^2}{2} = \boxed{106}.$

Solution 8 (Coordinates and algebraic manipulation)

$[asy] pair A,B,C,D,P; A=(-3,3); B=(3,3); C=(3,-3); D=(-3,-3); draw(A--B--C--D--cycle); label(A,"$A$",NW); label(B,"$B$",NE); label(C,"$C$",SE); label(D,"$D$",SW); draw(circle((0,0),4.24264068712)); P=(-1,4.12310562562); dot(P); label(P,"$P$", NW); pen k=red+dashed; draw(P--A,k); draw(P--B,k); draw(P--C,k); draw(P--D,k); [/asy]$ Let $P=(a,b)$ on the upper quarter of the circle, and let $k$ be the side length of the square. Hence, we want to find $k^2$ . Let the center of the circle be $(0,0)$ . The two equations would thus become: $\left(\left(a+\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a-\dfrac{k}2\right)^2+\left(b+\dfrac{k}2\right)^2\right)=56^2$ $\left(\left(a-\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a+\dfrac{k}2\right)^2+\left(b+\dfrac{k}2\right)^2\right)=90^2$ Now, let $m=\left(a+\dfrac{k}2\right)^2$ , $n=\left(a-\dfrac{k}2\right)^2$ , $o=\left(b+\dfrac{k}2\right)^2$ , and $p=\left(b-\dfrac{k}2\right)^2$ . Our equations now change to $(m+p)(n+o)=56^2=mn+op+mo+pn$ and $(n+p)(m+o)=90^2=mn+op+no+pm$ . Subtracting the first from the second, we have $pm+no-mo-pn=p(m-n)-o(m-n)=(m-n)(p-o)=34\cdot146$ . Substituting back in and expanding, we have $2ak\cdot-2bk=34\cdot146$ , so $abk^2=-17\cdot73$ . We now have one of our terms we need ( $k^2$ ). Therefore, we only need to find $ab$ to find $k^2$ . We now write the equation of the circle, which point $P$ satisfies: $a^2+b^2=\left(\dfrac{k\sqrt{2}}{2}\right)^2=\dfrac{k^2}2$ We can expand the second equation, yielding $\left(a^2+b^2+\dfrac{k^2}2+(ak+bk)\right)\left(a^2+b^2+\dfrac{k^2}2-(ak+bk)\right)=(k^2+k(a+b))(k^2-k(a+b))=8100.$ Now, with difference of squares, we get $k^4-k^2\cdot(a+b)^2=k^2(k^2-(a+b)^2)=8100$ . We can add $2abk^2=-17\cdot73\cdot2=-2482$ to this equation, which we can factor into $k^2(k^2-(a+b)^2+2ab)=k^2(k^2-(a^2+b^2))=8100-2482$ . We realize that $a^2+b^2$ is the same as the equation of the circle, so we plug its equation in: $k^2\left(k^2-\dfrac{k^2}2\right)=5618$ . We can combine like terms to get $k^2\cdot\dfrac{k^2}2=5618$ , so $(k^2)^2=11236$ . Since the answer is an integer, we know $11236$ is a perfect square. Since it is even, it is divisible by $4$ , so we can factor $11236=2^2\cdot2809$ . With some testing with approximations and last-digit methods, we can find that $53^2=2809$ . Therefore, taking the square root, we find that $k^2$ , the area of square $ABCD$ , is $2\cdot53=\boxed{106}$ .

2023 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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