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

Revision as of 09:45, 1 June 2023

1 Problem
2 Solution 1 (Ptolemy's Theorem)
3 Solution 2 (Areas and Pythagorean Theorem)
4 Solution 3 (Similar Triangles)
5 Solution 4 (Heights and Half-Angle Formula)
6 Solution 5 (Analytic Geometry)
7 Solution 6 (Law of Cosines)
8 Solution 7 (Subtended Chords)
9 Solution 8 (Coordinates and Algebraic Manipulation)
10 Solution 9 (Law of Sines)
11 Solution 10 (Areas and Trigonometry)
12 Solution 11 (it's just pythag, area and altitudes! 5 min sol)
13 Video Solution 1 by TheBeautyofMath
14 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 (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: $[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot("$A$", A, 1.5*NW, linewidth(4)); dot("$B$", B, 1.5*SW, linewidth(4)); dot("$C$", C, 1.5*SE, linewidth(4)); dot("$D$", D, 1.5*NE, linewidth(4)); dot("$P$", P, 1.5*dir(P), linewidth(4)); dot("$X$", X, 1.5*dir(20), linewidth(4)); dot("$Y$", Y, 1.5*dir(Y-P), linewidth(4)); dot("$O$", O, 1.5*E, linewidth(4)); [/asy]$ 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. By the Pythagorean Theorem, we have $PX^2+PY^2=PO^2.$ 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 $d^2=212.$ Therefore, we get $[ABCD] = \frac{d^2}{2} = \boxed{106}.$ ~MRENTHUSIASM

Solution 3 (Similar Triangles)

$[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot("$A$", A, 1.5*NW, linewidth(4)); dot("$B$", B, 1.5*SW, linewidth(4)); dot("$C$", C, 1.5*SE, linewidth(4)); dot("$D$", D, 1.5*NE, linewidth(4)); dot("$P$", P, 1.5*dir(P), linewidth(4)); dot("$X$", X, 1.5*dir(20), linewidth(4)); dot("$Y$", Y, 1.5*dir(Y-P), linewidth(4)); dot("$O$", O, 1.5*E, linewidth(4)); [/asy]$ 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 equation 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 4 (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 $\tan{\angle{POX}} = \frac{PX}{XO} = \frac{28}{45}$ by triangle area and given information. Then, notice that the measure of $\angle{OCP}$ is half of $\angle{XOP}$ .

Using the half-angle formula for tangent,

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

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 5 (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 6 (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 7 (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$", O, 1.5*S); label("$\theta$", O, dir(120)*5); draw(P--A--C--cycle, red); draw(P--B--D--cycle, blue); draw(P--O); draw(anglemark(P,O,A,30)); dot(P); dot(O); [/asy]$ 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 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$",SW); draw(circle((0,0),4.24264068712)); P=(-1,4.12310562562); label(P,"$P$", NW); pen k=red+dashed; draw(P--A,k); draw(P--B,k); draw(P--C,k); draw(P--D,k); dot(P); [/asy]$ 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 $2\cdot53=\boxed{106}$ .

~wuwang2002

Solution 9 (Law of Sines)

WLOG, let $P$ be on minor arc $AD.$ Draw in $AP$ , $BP$ , $CP$ , $DP$ and let $\angle ABP = x.$ We can see, by the inscribed angle theorem, that $\angle APB = \angle ACB = 45$ , and $\angle CPD = \angle CAD = 45.$ Then, $\angle PAB = 135-x$ , $\angle PCD = \angle PAD = (135-x)-90 = 45-x$ , and $\angle PDC = 90+x.$ Letting $(PA, PB, PC, PD, AB) = (a,b,c,d,s)$ , we can use the law of sines on triangles $PAB$ and $PCD$ to get $s\sqrt{2} = \frac{a}{\sin(x)} = \frac{b}{\sin(135-x)} = \frac{c}{\sin(90+x)} = \frac{d}{\sin(45-x)}.$ Making all the angles in the above equation acute gives $s\sqrt{2} = \frac{a}{\sin(x)} = \frac{b}{\sin(45+x)} = \frac{c}{\sin(90-x)} = \frac{d}{\sin(45-x)}.$

Note that we are looking for $s^{2}.$ We are given that $ac = 56$ and $bd = 90.$ This means that $s^{2}\sin(x)\sin(90-x) = 28$ and $s^{2}\sin(45+x)\sin(45-x) = 45.$ However, $\sin(x)\sin(90-x) = \sin(x)\cos(x) = \frac{\sin(2x)}{2}$ and $\sin(45+x)\sin(45-x) = \frac{(\cos(x) + \sin(x))(\cos(x) - \sin(x))}{2} = \frac{\cos^{2}(x) - \sin^{2}(x)}{2} = \frac{\cos(2x)}{2}.$ Therefore, $s^{2}\sin(2x) = 56$ and $s^{2}\cos(2x) = 90.$ Therefore, by the Pythagorean Identity, $s^{2} = \sqrt{(s^{2}\sin(2x))^{2} + (s^{2}\cos(2x))^{2}} = \sqrt{56^{2} + 90^{2}} = \boxed{106}.$

~pianoboy

Solution 10 (Areas and Trigonometry)

Similar to Solution 6, let $P$ be on minor arc $\overarc {AB}$ , $r$ and $O$ be the radius and center of the circumcircle respectively, and $\theta = \angle AOP$ . Since $\triangle APC$ is a right triangle, $PA \cdot PC$ equals the hypotenuse, $2r$ , times its altitude, which can be represented as $r \sin \theta$ . Therefore, $2r^2 \sin \theta = 56$ . Applying similar logic to $\triangle BPD$ , we get $2r^2 \sin (90^\circ - \theta) = 2r^2 \cos \theta = 90$ .

Dividing the two equations, we have $\begin{align*} \frac{\sin \theta}{\cos \theta} &= \frac{56}{90} \\ 56 \cos \theta &= 90 \sin \theta \\ (56 \cos \theta)^2 &= (90 \sin \theta)^2. \end{align*}$ Adding $(56 \sin \theta)^2$ to both sides allows us to get rid of $\cos \theta$ : $\begin{align*} (56 \cos \theta)^2 + (56 \sin \theta)^2 &= (90 \sin \theta)^2 + (56 \sin \theta)^2 \\ 56^2 &= (90^2 + 56^2)(\sin \theta)^2 \\ \frac{56^2}{90^2 + 56^2} &= (\sin \theta)^2 \\ \frac{28}{53} &= \sin \theta. \end{align*}$ Therefore, we have $2r^2\left(\frac{28}{53}\right) = 56$ , and since the area of the square can be represented as $2r^2$ , the answer is $56 \cdot \frac{53}{28} = \boxed{106}$ .

~phillipzeng

Solution 11 (it's just pythag, area and altitudes! 5 min sol)

$[asy] size(6cm,0); draw(circle((0,0),2)); pair O = (0,0); pair A = (-1.414,1.414); pair B = (1.414,1.414); pair C = (1.414,-1.414); pair D = (-1.414,-1.414); pair P = (0.3,1.977); dot(O); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(P--A,red); draw(P--B,blue); draw(P--C,red); draw(P--D,blue); label("$O$",O,N); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,N); [/asy]$

The key observation to make in this problem is that both $AC$ and $BD$ are diameters, so $\triangle APC$ and $\triangle PBD$ are right triangles. Thus, $AP \cdot PC = 56 = 2[APC] \longrightarrow [APC] = 28$ and $BP \cdot PD = 90 = 2[PBD] \longrightarrow [PBD] = 45$ . We thus are motivated to find a second way to express $[APC]$ and $[PBD]$ to solve for unknown side lengths. We do this by dropping altitudes from $P$ to $AC$ and $BD$ (denote the foot of these altitudes $E$ and $F$ , respectively):

$[asy] size(6cm,0); draw(circle((0,0),2)); pair O = (0,0); pair A = (-1.414,1.414); pair B = (1.414,1.414); pair C = (1.414,-1.414); pair D = (-1.414,-1.414); pair P = (0.3,1.977); pair E = (-0.839,0.839); pair F = (1.139,1.139); dot(O); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(P--A,red); draw(P--B,blue); draw(P--C,red); draw(P--D,blue); draw(A--C,red); draw(B--D,blue); draw(P--E,red); draw(P--F,blue); label("$O$",O,S); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,N); label("$E$",E,SW); label("$F$",F,SE); [/asy]$

To prevent cluttering of the diagram, we shall now remove $AP,CP,BP,$ and $DP$ from the picture:

$[asy] size(6cm,0); draw(circle((0,0),2)); pair O = (0,0); pair A = (-1.414,1.414); pair B = (1.414,1.414); pair C = (1.414,-1.414); pair D = (-1.414,-1.414); pair P = (0.3,1.977); pair E = (-0.839,0.839); pair F = (1.139,1.139); dot(O); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(A--C,red); draw(B--D,blue); draw(P--E,red); draw(P--F,blue); label("$O$",O,S); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,N); label("$E$",E,SW); label("$F$",F,SE); [/asy]$

Note that $AC \perp BD$ , so $\angle EOF = 90^\circ$ , $PE \perp AC$ , so $\angle PEO = 90^\circ$ , $PF \perp BD$ so $\angle PFO = 90^\circ$ , and finally $\angle EPF = 360 - 90 - 90 - 90 = 90^\circ$ . Thus, $EOFP$ is a rectangle. We know $[APC]=28=\dfrac{PE \cdot AC}{2}$ and $[PBD]=45=\dfrac{PF \cdot BD}{2}$ , and we also know that $AC=BD=2r$ since they're both diameters. Thus, $r \cdot PE = 28 \longrightarrow PE = \dfrac{28}{r}$ and $r \cdot PF = 45 \longrightarrow PF = \dfrac{45}{r}$ . We finish the problem by drawing in $PO$ :

$[asy] size(6cm,0); draw(circle((0,0),2)); pair O = (0,0); pair A = (-1.414,1.414); pair B = (1.414,1.414); pair C = (1.414,-1.414); pair D = (-1.414,-1.414); pair P = (0.3,1.977); pair E = (-0.839,0.839); pair F = (1.139,1.139); dot(O); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(A--C,red); draw(B--D,blue); draw(P--E,red); draw(P--F,blue); draw(P--O); label("$O$",O,S); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,N); label("$E$",E,SW); label("$F$",F,SE); [/asy]$

$\triangle EOP$ is right since $EOFP$ is a rectangle, and $PO=r$ since it's a radii. Thus, by Pythagorean theorem, we have that $\begin{alignat*}{8} PE^2+EO^2&=PO^2 \\ \left( \dfrac{28}{r} \right)^2 + \left( \dfrac{45}{r} \right)^2 &= r^2 \\ 28^2+45^2&=r^2 \\ 2809&=r^2 \\ r&=\sqrt{53}. \\ \end{alignat*}$

Finally, $AB=r\sqrt{2}$ since $AOB$ is a $45-45-90$ triangle with $AO=OB=r$ , so $[ABCD]=AB^2=2r^2$ . Our answer is thus $\boxed{106}$ .

Note: While this solution seems very long, in practice it only takes at most 3-5 minutes if done efficiently, as all the steps are extremely trivial and motivated and during the contest only one diagram needs to be drawn rather than the four used above for clarity.

~ Sleepy_Head

Video Solution 1 by TheBeautyofMath

https://youtu.be/JMxOWyF3i20

~IceMatrix

2023 AIME I (Problems • Answer Key • Resources)
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