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

(Solution 2 (Trigonometry))
(Solution 10 (Areas and Trigonometry))
 
(109 intermediate revisions by 17 users not shown)
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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):
+
==Problem==
  
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?
+
Let <math>P</math> be a point on the circle circumscribing square <math>ABCD</math> that satisfies <math>PA \cdot PC = 56</math> and <math>PB \cdot PD = 90.</math> Find the area of <math>ABCD.</math>
  
==Solution (Ptolemy's Theorem)==
+
==Solution 1 (Ptolemy's Theorem)==
  
 
[[Ptolemy's theorem]] states that for cyclic quadrilateral <math>WXYZ</math>, <math>WX\cdot YZ + XY\cdot WZ = WY\cdot XZ</math>.
 
[[Ptolemy's theorem]] states that for cyclic quadrilateral <math>WXYZ</math>, <math>WX\cdot YZ + XY\cdot WZ = WY\cdot XZ</math>.
  
We may assume that <math>P</math> is between <math>B</math> and <math>C</math>. Let <math>PA = a</math>, <math>PB = b</math>, <math>PC = C</math>, <math>PD = d</math>, and <math>AB = s</math>. We have <math>a^2 + c^2 = AC^2 = 2s^2</math>, because <math>AC</math> is a diameter of the circle. Similarly, <math>b^2 + d^2 = 2s^2</math>. Therefore, <math>(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112</math>. Similarly, <math>(b+d)^2 = 2s^2 + 180</math>.  
+
We may assume that <math>P</math> is between <math>B</math> and <math>C</math>. Let <math>PA = a</math>, <math>PB = b</math>, <math>PC = c</math>, <math>PD = d</math>, and <math>AB = s</math>. We have <math>a^2 + c^2 = AC^2 = 2s^2</math>, because <math>AC</math> is a diameter of the circle. Similarly, <math>b^2 + d^2 = 2s^2</math>. Therefore, <math>(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112</math>. Similarly, <math>(b+d)^2 = 2s^2 + 180</math>.  
  
 
By Ptolemy's Theorem on <math>PCDA</math>, <math>as + cs = ds\sqrt{2}</math>, and therefore <math>a + c = d\sqrt{2}</math>. By Ptolemy's on <math>PBAD</math>, <math>bs + ds = as\sqrt{2}</math>, and therefore <math>b + d = a\sqrt{2}</math>. By squaring both equations, we obtain
 
By Ptolemy's Theorem on <math>PCDA</math>, <math>as + cs = ds\sqrt{2}</math>, and therefore <math>a + c = d\sqrt{2}</math>. By Ptolemy's on <math>PBAD</math>, <math>bs + ds = as\sqrt{2}</math>, and therefore <math>b + d = a\sqrt{2}</math>. By squaring both equations, we obtain
 +
<cmath>\begin{alignat*}{8}
 +
2d^2 &= (a+c)^2 &&= 2s^2 + 112, \\
 +
2a^2 &= (b+d)^2 &&= 2s^2 + 180.
 +
\end{alignat*}</cmath>
 +
Thus, <math>a^2 = s^2 + 90</math>, and <math>d^2 = s^2 + 56</math>. Plugging these values into <math>a^2 + c^2 = b^2 + d^2 = 2s^2</math>, we obtain <math>c^2 = s^2 - 90</math>, and <math>b^2 = s^2 - 56</math>. Now, we can solve using <math>a</math> and <math>c</math> (though using <math>b</math> and <math>d</math> yields the same solution for <math>s</math>).
 +
<cmath>\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*}</cmath>
 +
~mathboy100
 +
 +
==Solution 2 (Areas and Pythagorean Theorem)==
 +
 +
By the <b>Inscribed Angle Theorem</b>, we conclude that <math>\triangle PAC</math> and <math>\triangle PBD</math> are right triangles.
 +
 +
Let the brackets denote areas. We are given that
 +
<cmath>\begin{alignat*}{8}
 +
2[PAC] &= PA \cdot PC &&= 56, \\
 +
2[PBD] &= PB \cdot PD &&= 90.
 +
\end{alignat*}</cmath>
 +
Let <math>O</math> be the center of the circle, <math>X</math> be the foot of the perpendicular from <math>P</math> to <math>\overline{AC},</math> and <math>Y</math> be the foot of the perpendicular from <math>P</math> to <math>\overline{BD},</math> as shown below:
 +
<asy>
 +
/* Made by MRENTHUSIASM */
  
<cmath>2d^2 = (a+c)^2 = 2s^2 + 112</cmath>
+
size(200);
<cmath>2a^2 = (b+d)^2 = 2s^2 + 180.</cmath>
+
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;
  
Thus, <math>a^2 = s^2 + 90</math>, and <math>d^2 = s^2 + 56</math>. Plugging these values into <math>a^2 + c^2 = b^2 + d^2 = 2s^2</math>, we obtain <math>c^2 = s^2 - 90</math>, and <math>b^2 = s^2 - 56</math>. Now, we can solve using <math>a</math> and <math>c</math> (though using <math>b</math> and <math>d</math> yields the same solution for <math>s</math>).
+
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 <math>d</math> be the diameter of <math>\odot O.</math> It follows that
 +
<cmath>\begin{alignat*}{8}
 +
2[PAC] &= d\cdot PX &&= 56, \\
 +
2[PBD] &= d\cdot PY &&= 90.
 +
\end{alignat*}</cmath>
 +
Moreover, note that <math>OXPY</math> is a rectangle. By the Pythagorean Theorem, we have <cmath>PX^2+PY^2=PO^2.</cmath>
 +
We rewrite this equation in terms of <math>d:</math> <cmath>\left(\frac{56}{d}\right)^2+\left(\frac{90}{d}\right)^2=\left(\frac d2\right)^2,</cmath> from which <math>d^2=212.</math> Therefore, we get <cmath>[ABCD] = \frac{d^2}{2} = \boxed{106}.</cmath>
 +
~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);
  
<cmath>(\sqrt{s^2 + 90})(\sqrt{s^2 - 90}) = ac = 56</cmath>
+
P = intersectionpoints(Circle(A,4),p)[1];
<cmath>(s^2 + 90)(s^2 - 90) = 56^2</cmath>
+
X = foot(P,A,C);
<cmath>s^4 = 90^2 + 56^2 = 106^2</cmath>
+
Y = foot(P,B,D);
<cmath>s^2 = 106.</cmath>
 
  
The answer is <math>\boxed{106}</math>.
+
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 <math>O</math>, and the radius of the circle be <math>r</math>. Since <math>ABCD</math> is a rhombus with diagonals <math>2r</math> and <math>2r</math>, its area is <math>\dfrac{1}{2}(2r)(2r) = 2r^2</math>. Since <math>AC</math> and <math>BD</math> are diameters of the circle, <math>\triangle APC</math> and <math>\triangle BPD</math> are right triangles. Let <math>X</math> and <math>Y</math> be the foot of the altitudes to <math>AC</math> and <math>BD</math>, respectively. We have
 +
<cmath>[\triangle APC] = \frac{1}{2}(PA)(PC) = \frac{1}{2}(PX)(AC),</cmath>
 +
so <math>PX = \dfrac{(PA)(PC)}{AC} = \dfrac{28}{r}</math>. Similarly,
 +
<cmath>[\triangle BPD] = \frac{1}{2}(PB)(PD) = \frac{1}{2}(PY)(PB),</cmath>
 +
so <math>PY = \dfrac{(PB)(PD)}{BD} = \dfrac{45}{r}</math>. Since <math>\triangle APX \sim \triangle PCX,</math>
 +
<cmath>\frac{AX}{PX} = \frac{PX}{CX}</cmath>
 +
<cmath>\frac{AO - XO}{PX} = \frac{PX}{OC + XO}.</cmath>
 +
But <math>PXOY</math> is a rectangle, so <math>PY = XO</math>, and our equation becomes
 +
<cmath>\frac{r - PY}{PX} = \frac{PX}{r + PY}.</cmath>
 +
Cross multiplying and rearranging gives us <math>r^2 = PX^2 + PY^2 = \left(\dfrac{28}{r}\right)^2 + \left(\dfrac{45}{r}\right)^2</math>, which rearranges to <math>r^4 = 2809</math>. Therefore <math>[ABCD] = 2r^2 = \boxed{106}</math>.
  
~mathboy100
+
~Cantalon
  
==Solution 2 (Trigonometry)==
+
==Solution 4 (Heights and Half-Angle Formula)==
Drop a height from point <math>P</math> to line <math>\overline{AC}</math> and line <math>\overline{BC}</math>. 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.
+
Drop a height from point <math>P</math> to line <math>\overline{AC}</math> and line <math>\overline{BC}</math>. Call these two points to be <math>X</math> and <math>Y</math>, respectively. Notice that the intersection of the diagonals of <math>\square ABCD</math> meets at a right angle at the center of the circumcircle, call this intersection point <math>O</math>.
  
Since OXPY is a rectangle, OX is the distance from P to line BD. We know 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 angle XOY.
+
Since <math>OXPY</math> is a rectangle, <math>OX</math> is the distance from <math>P</math> to line <math>\overline{BD}</math>. We know that <math>\tan{\angle{POX}} = \frac{PX}{XO} = \frac{28}{45}</math> by triangle area and given information. Then, notice that the measure of <math>\angle{OCP}</math> is half of <math>\angle{XOP}</math>.
  
 
Using the half-angle formula for tangent,
 
Using the half-angle formula for tangent,
Line 34: Line 141:
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
\frac{(2*\tan{\angle{OCP}})}{(1-\tan{\angle{OCP}}^2)} = \tan{\angle{YOX}} = \frac{28}{45}
+
\frac{(2 \cdot \tan{\angle{OCP}})}{(1-\tan^2{\angle{OCP}})} = \tan{\angle{POX}} = \frac{28}{45}
 
\\
 
\\
14\tan{\angle{OCP}}^2 + 45\tan{\angle{OCP}} - 14 = 0
+
14\tan^2{\angle{OCP}} + 45\tan{\angle{OCP}} - 14 = 0
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
  
we get that <math>\tan{\angle{OCP}} = -7/2</math> or <math>2/7</math>. Since this value must be positive, we pick 2/7. Then, PA/PC = 2/7 (since triangle CAP is a right triangle with AC also the diameter of the circumcircle) and PA * PC = 56. Solving we get PA = 4, PC = 14, giving us a diagonal of length <math>\sqrt{212}</math> and area <math>\boxed{106}</math>.
+
Solving the equation above, we get that <math>\tan{\angle{OCP}} = -7/2</math> or <math>2/7</math>. Since this value must be positive, we pick <math>\frac{2}{7}</math>. Then, <math>\frac{PA}{PC} = 2/7</math> (since <math>\triangle CAP</math> is a right triangle with line <math>\overline{AC}</math> the diameter of the circumcircle) and <math>PA * PC = 56</math>. Solving we get <math>PA = 4</math>, <math>PC = 14</math>, giving us a diagonal of length <math>\sqrt{212}</math> and area <math>\boxed{106}</math>.
  
~Danielzh
+
~[[Daniel Zhou's Profile|Danielzh]]
  
==Solution 3 (Analytic geometry)==
+
==Solution 5 (Analytic Geometry)==
  
 
Denote by <math>x</math> the half length of each side of the square.
 
Denote by <math>x</math> the half length of each side of the square.
Line 102: Line 209:
 
</cmath>
 
</cmath>
  
+
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
==Solution 4 (Law of Cosines)==
+
 
 +
==Solution 6 (Law of Cosines)==
 
WLOG, let <math>P</math> be on minor arc <math>\overarc {AB}</math>. Let <math>r</math> and <math>O</math> be the radius and center of the circumcircle respectively, and let <math>\theta = \angle AOP</math>. 
 
WLOG, let <math>P</math> be on minor arc <math>\overarc {AB}</math>. Let <math>r</math> and <math>O</math> be the radius and center of the circumcircle respectively, and let <math>\theta = \angle AOP</math>. 
 
 
Line 120: Line 228:
 
~OrangeQuail9
 
~OrangeQuail9
  
==Solution 5==
+
==Solution 7 (Subtended Chords)==
test
+
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 <math>r</math>. Then the area of the <math>ABCD</math> is <math>(\sqrt2r)^2 = 2r^2</math>
 +
Using the formula for the length of a chord subtended by an angle, we get
 +
<cmath>PA = 2r\sin\left(\dfrac{\theta}2\right)</cmath>
 +
<cmath>PC = 2r\sin\left(\dfrac{180-\theta}2\right) = 2r\sin\left(90 - \dfrac{\theta}2\right) = 2r\cos\left(\dfrac{\theta}2\right)</cmath>
 +
Multiplying and simplifying these 2 equations gives
 +
<cmath>PA \cdot PC = 4r^2 \sin \left(\dfrac{\theta}2 \right) \cos \left(\dfrac{\theta}2 \right) = 2r^2 \sin\left(\theta \right) = 56</cmath>
 +
Similarly <math>PB = 2r\sin\left(\dfrac{90 +\theta}2\right)</math> and <math>PD =2r\sin\left(\dfrac{90 -\theta}2\right)</math>. Again, multiplying gives
 +
<cmath>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)</cmath>
 +
<cmath> =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</cmath>
 +
Dividing <math>2r^2 \sin \left(\theta \right)</math> by <math>2r^2 \cos \left( \theta \right)</math> gives <math>\tan \left(\theta \right) = \dfrac{28}{45}</math>, so <math>\theta = \tan^{-1} \left(\dfrac{28}{45} \right)</math>.
 +
Pluging this back into one of the equations, gives
 +
<cmath>2r^2 = \dfrac{90}{\cos\left(\tan^{-1}\left(\dfrac{28}{45}\right)\right)}</cmath>
 +
If we imagine a <math>28</math>-<math>45</math>-<math>53</math> right triangle, we see that if <math>28</math> is opposite and <math>45</math> is adjacent, <math>\cos\left(\theta\right) = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{45}{53}</math>. Now we see that
 +
<cmath>2r^2 = \dfrac{90}{\frac{45}{53}} = \boxed{106}.</cmath>
 +
~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 <math>P=(a,b)</math> on the upper quarter of the circle, and let <math>k</math> be the side length of the square. Hence, we want to find <math>k^2</math>. Let the center of the circle be <math>(0,0)</math>.
 +
The two equations would thus become:
 +
<cmath>\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</cmath>
 +
<cmath>\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</cmath>
 +
Now, let <math>m=\left(a+\dfrac{k}2\right)^2</math>, <math>n=\left(a-\dfrac{k}2\right)^2</math>, <math>o=\left(b+\dfrac{k}2\right)^2</math>, and <math>p=\left(b-\dfrac{k}2\right)^2</math>. Our equations now change to <math>(m+p)(n+o)=56^2=mn+op+mo+pn</math> and <math>(n+p)(m+o)=90^2=mn+op+no+pm</math>. Subtracting the first from the second, we have <math>pm+no-mo-pn=p(m-n)-o(m-n)=(m-n)(p-o)=34\cdot146</math>. Substituting back in and expanding, we have <math>2ak\cdot-2bk=34\cdot146</math>, so <math>abk^2=-17\cdot73</math>. We now have one of our terms we need (<math>k^2</math>). Therefore, we only need to find <math>ab</math> to find <math>k^2</math>.
 +
We now write the equation of the circle, which point <math>P</math> satisfies: <cmath>a^2+b^2=\left(\dfrac{k\sqrt{2}}{2}\right)^2=\dfrac{k^2}2</cmath>
 +
We can expand the second equation, yielding <cmath>\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.</cmath>
 +
Now, with difference of squares, we get <math>k^4-k^2\cdot(a+b)^2=k^2(k^2-(a+b)^2)=8100</math>. We can add <math>2abk^2=-17\cdot73\cdot2=-2482</math> to this equation, which we can factor into <math>k^2(k^2-(a+b)^2+2ab)=k^2(k^2-(a^2+b^2))=8100-2482</math>. We realize that <math>a^2+b^2</math> is the same as the equation of the circle, so we plug its equation in: <math>k^2\left(k^2-\dfrac{k^2}2\right)=5618</math>. We can combine like terms to get <math>k^2\cdot\dfrac{k^2}2=5618</math>, so <math>(k^2)^2=11236</math>.
 +
Since the answer is an integer, we know <math>11236</math> is a perfect square. Since it is even, it is divisible by <math>4</math>, so we can factor <math>11236=2^2\cdot2809</math>. With some testing with approximations and last-digit methods, we can find that <math>53^2=2809</math>. Therefore, taking the square root, we find that <math>k^2</math>, the area of square <math>ABCD</math>, is <math>2\cdot53=\boxed{106}</math>.
 +
 
 +
~wuwang2002
 +
 
 +
==Solution 9 (Law of Sines)==
 +
 
 +
WLOG, let <math>P</math> be on minor arc <math>AD.</math> Draw in <math>AP</math>, <math>BP</math>, <math>CP</math>, <math>DP</math> and let <math>\angle ABP = x.</math>  We can see, by the inscribed angle theorem, that <math>\angle APB = \angle ACB = 45</math>, and <math>\angle CPD = \angle CAD = 45.</math> Then, <math>\angle PAB = 135-x</math>, <math>\angle PCD = \angle PAD = (135-x)-90 = 45-x</math>, and <math>\angle PDC = 90+x.</math> Letting <math>(PA, PB, PC, PD, AB) = (a,b,c,d,s)</math>, we can use the law of sines on triangles <math>PAB</math> and <math>PCD</math> to get <cmath>s\sqrt{2} = \frac{a}{\sin(x)} = \frac{b}{\sin(135-x)} = \frac{c}{\sin(90+x)} = \frac{d}{\sin(45-x)}.</cmath> Making all the angles in the above equation acute gives <cmath>s\sqrt{2} = \frac{a}{\sin(x)} = \frac{b}{\sin(45+x)} = \frac{c}{\sin(90-x)} = \frac{d}{\sin(45-x)}.</cmath>
 +
 
 +
Note that we are looking for <math>s^{2}.</math> We are given that <math>ac = 56</math> and <math>bd = 90.</math> This means that
 +
<math>s^{2}\sin(x)\sin(90-x) = 28</math> and <math>s^{2}\sin(45+x)\sin(45-x) = 45.</math> However, <cmath>\sin(x)\sin(90-x) = \sin(x)\cos(x) = \frac{\sin(2x)}{2}</cmath> and <cmath>\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}.</cmath>  Therefore, <math>s^{2}\sin(2x) = 56</math> and <math>s^{2}\cos(2x) = 90.</math> Therefore, by the Pythagorean Identity, <cmath>s^{2} = \sqrt{(s^{2}\sin(2x))^{2} + (s^{2}\cos(2x))^{2}} = \sqrt{56^{2} + 90^{2}} = \boxed{106}.</cmath>
 +
 
 +
~pianoboy
 +
 
 +
==Solution 10 (Areas and Trigonometry)==
 +
 
 +
Similar to Solution 6, let <math>P</math> be on minor arc <math>\overarc {AB}</math>, <math>r</math> and <math>O</math> be the radius and center of the circumcircle respectively, and <math>\theta = \angle AOP</math>. Since <math>\triangle APC</math> is a right triangle, <math>PA \cdot PC</math> equals the hypotenuse, <math>2r</math>, times its altitude, which can be represented as <math>r \sin \theta</math>. Therefore, <math>2r^2 \sin \theta = 56</math>. Applying similar logic to <math>\triangle BPD</math>, we get <math>2r^2 \sin (90^\circ - \theta) = 2r^2 \cos \theta = 90</math>.
 +
 
 +
Dividing the two equations, we have
 +
<cmath>\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*}</cmath>
 +
Adding <math>(56 \sin \theta)^2</math> to both sides allows us to get rid of <math>\cos \theta</math>:
 +
<cmath>\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*}</cmath>
 +
Therefore, we have <math>2r^2\left(\frac{28}{53}\right) = 56</math>, and since the area of the square can be represented as <math>2r^2</math>, the answer is <math>56 \cdot \frac{53}{28} = \boxed{106}</math>.
 +
 
 +
~phillipzeng
 +
 
 +
 
 +
==Solution 11 (Angle Chasing and Trigonometric Identities)==
 +
First, we define a few points. Let <math>O</math> be the center of the circle, let <math>E</math> be the intersection of diameter <math>AC</math> and chord <math>PD</math>, and let <math>F</math> be the intersection of diameter <math>BD</math> and chord <math>PC</math>. We know that <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> are four corners of a square. Therefore, the arcs <math>AD</math>, <math>DC</math>, and <math>CB</math> are all <math>90</math> degrees. By inscribed angles, angle <math>APD</math>, angle <math>DPC</math>, and angle <math>CPB</math> are <math>45</math> degrees each. Let the measure of angle <math>PAC</math> be <math>a</math>. Similarly, let the measure of angle <math>PBD</math> be <math>b</math>.
 +
 
 +
 
 +
Angle chasing will lead us to the fact that <math>a + b = 135</math>, or rather, <math>b = 135-a</math>. Let the diameter of the circle be <math>d</math>. Given by the problem, <math>d^2\sin a \cos a = 56</math>. Also, <math>d^2\sin b \cos b = 90</math>. Using the trigonometric identity <math>\sin 2x = 2\sin x \cos x</math>, we can rewrite these as <math>d^2\sin 2a = 112</math> and <math>d^2\sin 2b = 180</math>. Since we determined that <math>b = \frac{3\pi}{4}-a</math>, this can be substituted into the second equation. Then, we divide the two equations to get <math>\frac{\sin (\frac{3\pi}{2}-2a)}{\sin 2a} = \frac{45}{28}</math>. By using the trigonometric difference-of-angle identity, this simplifies to <math>\frac{-\cos 2a}{\sin 2a} = \frac{45}{28}</math>. By the definition of the tangent function, <math>\tan 2a = -\frac{28}{45}</math>
 +
 
 +
 
 +
Considering this hypothetical right triangle with legs of <math>28</math> and <math>45</math>, the hypotenuse is <math>\sqrt{45^2+28^2} = 53</math>. Since <math>\sin 2a</math> must be positive (since <math>a</math> is acute), <math>\sin 2a = \frac{28}{53}</math>. Substituting this into the first of the equations, <math>\frac{28}{53}d^2 = 112</math>. From this, <math>d^2 = 212</math>. The area of square <math>ABCD</math> is half of the square of its diagonal, which is <math>d</math>. Thus, the answer is <math>\frac{d^2}{2} = \boxed{106}</math>.
 +
 
 +
~Curious_crow
 +
 
 +
==Video Solution 1 by TheBeautyofMath==
 +
https://youtu.be/JMxOWyF3i20
 +
 
 +
~IceMatrix
 +
 
 +
==See also==
 +
{{AIME box|year=2023|num-b=4|num-a=6|n=I}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:33, 1 August 2024

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 (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

Video Solution 1 by TheBeautyofMath

https://youtu.be/JMxOWyF3i20

~IceMatrix

See also

2023 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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