Difference between revisions of "2016 AMC 10A Problems/Problem 24"

Revision as of 10:14, 10 April 2016

Problem

A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?

$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$

Solution

Solution 1 (Algebra)

To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.

$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(B,D,O,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$E$",E,WSW); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(B--D); draw(rightanglemark(C,E,D)); [/asy]$

Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Let the intersection of $BD$ and $OC$ be point $E$ . Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.

We set lengths $BE=ED$ equal to $x$ . By the Pythagorean Theorem, $\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}$

We solve for $x$ : $1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2$ $2\sqrt{(1-x^2)(2-x^2)}=2x^2-1$ $4(1-x^2)(2-x^2)=(2x^2-1)^2$ $8-12x^2+4x^4=4x^4-4x^2+1$ $8x^2=7$ $x=\frac{\sqrt{14}}{4}$

By Ptolemy's Theorem, $AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2$

Substituting values, $1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2$ $1+AD=\frac{7}{2}$ $AD=\frac{5}{2}$

Finally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\boxed{500 (E)}$ .

Solution 2 (Trigonometry Bash)

$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); [/asy]$

Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Apply law of cosines on $\Delta BOC$ ; let $\theta = \angle BOC$ . We get the following equation: $(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta$ Substituting the values in, we get $(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta$ Canceling out, we get $\cos\theta=\frac{3}{4}$ Because $\angle AOB$ , $\angle BOC$ , and $\angle COD$ are congruent, $\angle AOD = 3\theta$ . To find the remaining side ( $AD$ ), we simply have to apply the law of cosines to $\Delta AOD$ . Now, to find $\cos 3\theta$ , we can derive a formula that only uses $\cos\theta$ : $\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- \sin 2\theta \cdot (2\sin\theta \cos\theta)$ $\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-1+2\cos^{2}\theta)$ $\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta$ Plugging in $\cos\theta=\frac{3}{4}$ , we get $\cos 3\theta= -\frac{9}{16}$ . Now, applying law of cosines on triangle $OAD$ , we get $(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}$ $\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}$ $AD=200 \cdot \frac{5}{2}=\boxed{500}$

Solution 3 (Easier trig)

$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(A--E); draw(E--B); draw(C--F); draw(F--D); label("$E$",E,NW); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); draw(rightanglemark(A,E,B)); draw(rightanglemark(C,F,D)); [/asy]$

Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the missing side. Then, drop perpendiculars from $A$ and $D$ to (extended) line $BC$ , and let these points be $E$ and $F$ , respectively. Also, let $\theta = \angle BOC$ . From Law of Cosines on $\triangle BOC$ , we have $\cos \theta = \frac{3}{4}$ . Now, since $\triangle BOC$ is isosceles with $OB = OC$ , we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$ . By SSS congruence, we have that $\triangle OBC \cong \triangle OCD$ , so we have that $\angle OCD = \angle BCO = 90 - \frac{\theta}{2}$ , so $\angle DCF = \theta$ . Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$ , so $FC = 150$ . Similarly, $BE = 150$ , and $AD = 150 + 200 + 150 = \boxed{500}$ .

Solution 4 (Just Geometry)

$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label("$E$",E,NE); F=extension(C,O,A,D); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$

Label AD intercept OB at E and OC at F.

$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$

$\angle{BAD}=\frac{1}{2} \bullet \overarc{BCD}=\theta=\angle{AOB}$

so, $\triangle{OAB} \sim \triangle{ABE}$

we can get $AE=AB$ , similar $DF=CD$

$\frac{BE}{AB}=\frac{AB}{OA}$ , we could get $BE=\frac{AB^2}{OA}=\frac{200^2}{200\sqrt{2}}=100\sqrt{2}$

$OE = 100\sqrt{2} = \frac{1}{2}OB$ , $EF=\frac{1}{2}*BC=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{500}$

Solution 5 (Ptolemy's Theorem)

$[asy] pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D("A", dir(50), dir(50)); pair B = D("B", dir(90), dir(90)); pair C = D("C", dir(130), dir(130)); pair D = D("D", dir(170), dir(170)); pair O = D("O", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); [/asy]$

Let $s = 200$ . Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ . Since $\triangle OBC$ is isosceles we can compute its area to be $s^2 \sqrt7/4$ , hence $CA = 2 \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{7/2}$ .

Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = (7/2-1)s.$ This gives us: $\boxed{\textbf{(E) } 500.}$

Solution 6

Using trig. Since all three sides equal $200$ , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\sqrt{7},200\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}$ . Similarly, the cosine is $\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}$ . Since there are three sides, and since $\sin\theta=\sin\left(180-\theta\right)$ ,we seek to find $2r\sin 3\theta$ . First, $\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}$ and $\cos 2\theta=\frac{3}{4}$ by Pythagorean. $\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}$ $2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(E)}\text{ 500}}$

Solution 7 (Area)

For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$ , where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$ . If $M$ and $N$ are the midpoints of $BC$ and $AD$ , respectively, the height of the trapezoid is $OM-ON$ . By the pythagorean theorem, $OM=\sqrt{OB^2-BM^2}=\sqrt7$ and $ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}$ . Thus the height of the trapezoid is $\sqrt7-\sqrt{8-x^2}$ , so the area is $\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})$ . By Brahmagupta's formula, the area is $\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Setting these two equal, we get $(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Dividing both sides by $x+1$ and then squaring, we get $7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)$ . Expanding the right hand side and canceling the $x^2$ terms gives us $15-2(\sqrt7)(\sqrt{8-x^2}=2x+3$ . Rearranging and dividing by two, we get $(\sqrt7)(\sqrt{8-x^2})=6-x$ . Squaring both sides, we get $56-7x^2=x^2-12x+36$ . Rearranging, we get $8x^2-12x-20=0$ . Dividing by 4 we get $2x^2-3x-5=0$ . Factoring we get, $(2x-5)(x+1)=0$ , and since x cannot be negative, we get $x=2.5$ . Since $DA=2x$ , $DA=5$ . Scaling up by 100, we get $\boxed{\textbf{(E)}\text{ 500}}$ .

@@ Line 265: / Line 265: @@
 <cmath>\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}</cmath>
 <cmath>2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(E)}\text{ 500}}</cmath>
+===Solution 7 (Area)===
+For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=2</math> and <math>DA</math> is the missing side length. Let <math>DA=2x</math>. If <math>M</math> and <math>N</math> are the midpoints of <math>BC</math> and <math>AD</math>, respectively, the height of the trapezoid is <math>OM-ON</math>. By the pythagorean theorem, <math>OM=\sqrt{OB^2-BM^2}=\sqrt7</math> and <math>ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}</math>. Thus the height of the trapezoid is <math>\sqrt7-\sqrt{8-x^2}</math>, so the area is <math>\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})</math>. By Brahmagupta's formula, the area is <math>\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Setting these two equal, we get <math>(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Dividing both sides by <math>x+1</math> and then squaring, we get <math>7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)</math>. Expanding the right hand side and canceling the <math>x^2</math> terms gives us <math>15-2(\sqrt7)(\sqrt{8-x^2}=2x+3</math>. Rearranging and dividing by two, we get <math>(\sqrt7)(\sqrt{8-x^2})=6-x</math>. Squaring both sides, we get <math>56-7x^2=x^2-12x+36</math>. Rearranging, we get <math>8x^2-12x-20=0</math>. Dividing by 4 we get <math>2x^2-3x-5=0</math>. Factoring we get, <math>(2x-5)(x+1)=0</math>, and since x cannot be negative, we get <math>x=2.5</math>. Since <math>DA=2x</math>, <math>DA=5</math>. Scaling up by 100, we get <math>\boxed{\textbf{(E)}\text{ 500}}</math>.
 ==See Also==

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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