AIME 2023 I · 第 5 题

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 $s$).

$$\begin{aligned} ac = (\sqrt{s^2 - 90})(\sqrt{s^2 + 90}) &= 56 \\ (s^2 + 90)(s^2 - 90) &= 56^2 \\ s^4 &= 90^2 + 56^2 = 106^2 \\ s^2 &= \boxed{106}. \end{aligned}$$ ~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:

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)

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,

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

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

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

$$\begin{aligned} 2 x^4 \left( 1 - 2 \cos \theta \sin \theta \right) & = 28^2 \hspace{1cm} (1) \\ 2 x^4 \left( 1 + 2 \cos \theta \sin \theta \right) & = 45^2 \hspace{1cm} (2) \end{aligned}$$ 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

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

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

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)

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

$$\begin{aligned} \frac{\sin \theta}{\cos \theta} &= \frac{56}{90} \\ 56 \cos \theta &= 90 \sin \theta \\ (56 \cos \theta)^2 &= (90 \sin \theta)^2. \end{aligned}$$ Adding $(56 \sin \theta)^2$ to both sides allows us to get rid of $\cos \theta$:

$$\begin{aligned} (56 \cos \theta)^2 + (56 \sin \theta)^2 &= (90 \sin \theta)^2 + (56 \sin \theta)^2 \\ 56^2 &= (90^2 + 56^2)(\sin \theta)^2 \\ \frac{56^2}{90^2 + 56^2} &= (\sin \theta)^2 \\ \frac{28}{53} &= \sin \theta. \end{aligned}$$ 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 (Angle Chasing and Trigonometric Identities)

First, we define a few points. Let $O$ be the center of the circle, let $E$ be the intersection of diameter $AC$ and chord $PD$, and let $F$ be the intersection of diameter $BD$ and chord $PC$. We know that $A$, $B$, $C$, and $D$ are four corners of a square. Therefore, the arcs $AD$, $DC$, and $CB$ are all $90$ degrees. By inscribed angles, angle $APD$, angle $DPC$, and angle $CPB$ are $45$ degrees each. Let the measure of angle $PAC$ be $a$. Similarly, let the measure of angle $PBD$ be $b$.

Angle chasing will lead us to the fact that $a + b = 135$, or rather, $b = 135-a$. Let the diameter of the circle be $d$. Given by the problem, $d^2\sin a \cos a = 56$. Also, $d^2\sin b \cos b = 90$. Using the trigonometric identity $\sin 2x = 2\sin x \cos x$, we can rewrite these as $d^2\sin 2a = 112$ and $d^2\sin 2b = 180$. Since we determined that $b = \frac{3\pi}{4}-a$, this can be substituted into the second equation. Then, we divide the two equations to get $\frac{\sin (\frac{3\pi}{2}-2a)}{\sin 2a} = \frac{45}{28}$. By using the trigonometric difference-of-angle identity, this simplifies to $\frac{-\cos 2a}{\sin 2a} = \frac{45}{28}$. By the definition of the tangent function, $\tan 2a = -\frac{28}{45}$

Considering this hypothetical right triangle with legs of $28$ and $45$, the hypotenuse is $\sqrt{45^2+28^2} = 53$. Since $\sin 2a$ must be positive (since $a$ is acute), $\sin 2a = \frac{28}{53}$. Substituting this into the first of the equations, $\frac{28}{53}d^2 = 112$. From this, $d^2 = 212$. The area of square $ABCD$ is half of the square of its diagonal, which is $d$. Thus, the answer is $\frac{d^2}{2} = \boxed{106}$.

~Curious_crow

Solution 12 (Six Consecutive Equations)

This solution is like solution 1, but with more algebra than actual geometry.

Firstly, like in solution 1, label the respective lengths $PA$, $PB$, $PC$, $PD$, $AB = BC = CD = DA$ as $a$, $b$, $c$, $d$, $s$.

While only five equations is required, we use six to apply the shortcut of solving multistep equations, elimination.

Our six equations are as desired.

\begin{align} (5.1)\quad & ac = 56 && \text{Given} \\ (5.2)\quad & bd = 90 && \text{Given} \\ (5.3)\quad & b^2 + d^2 = 2s^2 && \text{Pythagorean on } \triangle BPD \\ (5.4)\quad & a^2 + c^2 = 2s^2 && \text{Pythagorean on } \triangle APC \\ (5.5)\quad & bs + as\sqrt{2} = ds && \text{Ptolemy on } APBD \\ (5.6)\quad & bs + ds = cs\sqrt{2} && \text{Ptolemy on } CBPD \end{align}

Note that rearranging $(5.6)$ gives $(5.6R)~bs - cs\sqrt{2} = -ds$ and thus $(5.5) + (5.6R)$ results in $2bs + as\sqrt{2} - cs\sqrt{2} = 0$, or $2b + \sqrt{2}(a-c) = 0$, and thus $b = \frac{\sqrt{2}(c-a)}{2}$. It then becomes obvious that $b$'s companion $d$ is $d = \frac{\sqrt{2}(c+a)}{2}$. Thus $(5.2)$ can be arranged to become $\frac{1}{2} \cdot (c^2 - a^2) = 90$, and thus $c^2 - a^2 = 180$. We have that $ab = 56$ and $c^2 - a^2 = 180$, in which we can then find $a$ and $c$ to be $4$ and $14$ (discard negative solutions as negative lengths do not exist in Plane Euclidean Geometry) respectively. Thus, we have for our final equation $a^2 + c^2 = 2s^2$, and we just need the result $s^2$ to solve the problem. Then our answer is simply $\frac{196+16}{2} = \frac{212}{2} = \boxed{106}$. $\square$

~Pinotation

Video Solution 1 by TheBeautyofMath

https://youtu.be/JMxOWyF3i20

~IceMatrix