Difference between revisions of "2023 AIME I Problems/Problem 5"
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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>). | 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*} | <cmath>\begin{align*} | ||
− | ac = (\sqrt{s^2 | + | ac = (\sqrt{s^2 - 90})(\sqrt{s^2 + 90}) &= 56 \\ |
(s^2 + 90)(s^2 - 90) &= 56^2 \\ | (s^2 + 90)(s^2 - 90) &= 56^2 \\ | ||
s^4 &= 90^2 + 56^2 = 106^2 \\ | s^4 &= 90^2 + 56^2 = 106^2 \\ | ||
Line 23: | Line 23: | ||
~mathboy100 | ~mathboy100 | ||
− | ==Solution 2 (Heights and Half-Angle Formula)== | + | ==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 */ | ||
+ | |||
+ | 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 <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); | ||
+ | |||
+ | 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 <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>. | ||
+ | |||
+ | ~Cantalon | ||
+ | |||
+ | ==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 <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>. | 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 <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{ | + | 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 32: | Line 141: | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \frac{(2 | + | \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 | 14\tan^2{\angle{OCP}} + 45\tan{\angle{OCP}} - 14 = 0 | ||
Line 40: | Line 149: | ||
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>. | 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 | + | ==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 211: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution | + | ==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 118: | Line 227: | ||
<cmath>2r^2 = \sqrt{56^2+90^2} = 2\sqrt{28^2+45^2} = 2\sqrt{2809} = 2 \cdot 53 = \boxed{106}.</cmath> | <cmath>2r^2 = \sqrt{56^2+90^2} = 2\sqrt{28^2+45^2} = 2\sqrt{2809} = 2 \cdot 53 = \boxed{106}.</cmath> | ||
~OrangeQuail9 | ~OrangeQuail9 | ||
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==Solution 7 (Subtended Chords)== | ==Solution 7 (Subtended Chords)== | ||
Line 306: | Line 303: | ||
~wuwang2002 | ~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== | ==See also== |
Latest revision as of 19:33, 1 August 2024
Contents
- 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 (Angle Chasing and Trigonometric Identities)
- 13 Video Solution 1 by TheBeautyofMath
- 14 See also
Problem
Let be a point on the circle circumscribing square
that satisfies
and
Find the area of
Solution 1 (Ptolemy's Theorem)
Ptolemy's theorem states that for cyclic quadrilateral ,
.
We may assume that is between
and
. Let
,
,
,
, and
. We have
, because
is a diameter of the circle. Similarly,
. Therefore,
. Similarly,
.
By Ptolemy's Theorem on ,
, and therefore
. By Ptolemy's on
,
, and therefore
. By squaring both equations, we obtain
Thus,
, and
. Plugging these values into
, we obtain
, and
. Now, we can solve using
and
(though using
and
yields the same solution for
).
~mathboy100
Solution 2 (Areas and Pythagorean Theorem)
By the Inscribed Angle Theorem, we conclude that and
are right triangles.
Let the brackets denote areas. We are given that
Let
be the center of the circle,
be the foot of the perpendicular from
to
and
be the foot of the perpendicular from
to
as shown below:
Let
be the diameter of
It follows that
Moreover, note that
is a rectangle. By the Pythagorean Theorem, we have
We rewrite this equation in terms of
from which
Therefore, we get
~MRENTHUSIASM
Solution 3 (Similar Triangles)
Let the center of the circle be
, and the radius of the circle be
. Since
is a rhombus with diagonals
and
, its area is
. Since
and
are diameters of the circle,
and
are right triangles. Let
and
be the foot of the altitudes to
and
, respectively. We have
so
. Similarly,
so
. Since
But
is a rectangle, so
, and our equation becomes
Cross multiplying and rearranging gives us
, which rearranges to
. Therefore
.
~Cantalon
Solution 4 (Heights and Half-Angle Formula)
Drop a height from point to line
and line
. Call these two points to be
and
, respectively. Notice that the intersection of the diagonals of
meets at a right angle at the center of the circumcircle, call this intersection point
.
Since is a rectangle,
is the distance from
to line
. We know that
by triangle area and given information. Then, notice that the measure of
is half of
.
Using the half-angle formula for tangent,
Solving the equation above, we get that or
. Since this value must be positive, we pick
. Then,
(since
is a right triangle with line
the diameter of the circumcircle) and
. Solving we get
,
, giving us a diagonal of length
and area
.
Solution 5 (Analytic Geometry)
Denote by the half length of each side of the square.
We put the square to the coordinate plane, with
,
,
,
.
The radius of the circumcircle of is
.
Denote by
the argument of point
on the circle.
Thus, the coordinates of
are
.
Thus, the equations and
can be written as
These equations can be reformulated as
These equations can be reformulated as
Taking , by solving the equation, we get
Plugging (3) into (1), we get
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 6 (Law of Cosines)
WLOG, let be on minor arc
. Let
and
be the radius and center of the circumcircle respectively, and let
.
By the Pythagorean Theorem, the area of the square is . We can use the Law of Cosines on isosceles triangles
to get
Taking the products of the first two and last two equations, respectively, and
Adding these equations,
so
~OrangeQuail9
Solution 7 (Subtended Chords)
First draw a diagram.
Let's say that the radius is
. Then the area of the
is
Using the formula for the length of a chord subtended by an angle, we get
Multiplying and simplifying these 2 equations gives
Similarly
and
. Again, multiplying gives
Dividing
by
gives
, so
.
Pluging this back into one of the equations, gives
If we imagine a
-
-
right triangle, we see that if
is opposite and
is adjacent,
. Now we see that
~Voldemort101
Solution 8 (Coordinates and Algebraic Manipulation)
Let
on the upper quarter of the circle, and let
be the side length of the square. Hence, we want to find
. Let the center of the circle be
.
The two equations would thus become:
Now, let
,
,
, and
. Our equations now change to
and
. Subtracting the first from the second, we have
. Substituting back in and expanding, we have
, so
. We now have one of our terms we need (
). Therefore, we only need to find
to find
.
We now write the equation of the circle, which point
satisfies:
We can expand the second equation, yielding
Now, with difference of squares, we get
. We can add
to this equation, which we can factor into
. We realize that
is the same as the equation of the circle, so we plug its equation in:
. We can combine like terms to get
, so
.
Since the answer is an integer, we know
is a perfect square. Since it is even, it is divisible by
, so we can factor
. With some testing with approximations and last-digit methods, we can find that
. Therefore, taking the square root, we find that
, the area of square
, is
.
~wuwang2002
Solution 9 (Law of Sines)
WLOG, let be on minor arc
Draw in
,
,
,
and let
We can see, by the inscribed angle theorem, that
, and
Then,
,
, and
Letting
, we can use the law of sines on triangles
and
to get
Making all the angles in the above equation acute gives
Note that we are looking for We are given that
and
This means that
and
However,
and
Therefore,
and
Therefore, by the Pythagorean Identity,
~pianoboy
Solution 10 (Areas and Trigonometry)
Similar to Solution 6, let be on minor arc
,
and
be the radius and center of the circumcircle respectively, and
. Since
is a right triangle,
equals the hypotenuse,
, times its altitude, which can be represented as
. Therefore,
. Applying similar logic to
, we get
.
Dividing the two equations, we have
Adding
to both sides allows us to get rid of
:
Therefore, we have
, and since the area of the square can be represented as
, the answer is
.
~phillipzeng
Solution 11 (Angle Chasing and Trigonometric Identities)
First, we define a few points. Let be the center of the circle, let
be the intersection of diameter
and chord
, and let
be the intersection of diameter
and chord
. We know that
,
,
, and
are four corners of a square. Therefore, the arcs
,
, and
are all
degrees. By inscribed angles, angle
, angle
, and angle
are
degrees each. Let the measure of angle
be
. Similarly, let the measure of angle
be
.
Angle chasing will lead us to the fact that , or rather,
. Let the diameter of the circle be
. Given by the problem,
. Also,
. Using the trigonometric identity
, we can rewrite these as
and
. Since we determined that
, this can be substituted into the second equation. Then, we divide the two equations to get
. By using the trigonometric difference-of-angle identity, this simplifies to
. By the definition of the tangent function,
Considering this hypothetical right triangle with legs of and
, the hypotenuse is
. Since
must be positive (since
is acute),
. Substituting this into the first of the equations,
. From this,
. The area of square
is half of the square of its diagonal, which is
. Thus, the answer is
.
~Curious_crow
Video Solution 1 by TheBeautyofMath
~IceMatrix
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.