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

Revision as of 17:16, 16 November 2022

The following problem is from both the 2016 AMC 10A #24 and 2016 AMC 12A #21, so both problems redirect to this page.

1 Problem
2 Solution 1 (Just Geometry)
3 Solution 2 (Algebra)
4 Solution 3 (HARD Algebra)
5 Solution 4 (Trigonometry Bash)
6 Solution 5 (Easier Trigonometry)
7 Solution 6 (Ptolemy's Theorem)
8 Solution 7 (Trigonometry)
9 Solution 8 (Area By Brahmagupta's Formula)
10 Solution 9 (Similar Triangles)
11 Solution 10 (Parameshwara’s Formula for Circumradius)
12 Solution 11 (Complex Numbers)
13 Video Solution by AoPS (Deven Ware)
14 Video Solution by Walt S.
15 Video Solution (Ptolemy’s Theorem)
16 Video Solution by TheBeautyofMath
17 Video Solution by Punxsutawney Phil
18 See Also

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 1 (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$",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$

Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$

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

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

From there, $\triangle{OAB} \sim \triangle{ABE}$ , thus:

$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$

$OA = OB$ because they are both radii of $\odot{O}$ . Since $\frac{OA}{AB} = \frac{OB}{AE}$ , we have that $AB = AE$ . Similarly, $CD = DF$ .

$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}$

Solution 2 (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 $x$ (Solution 1.1 begins from here). 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 3 (HARD Algebra)

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

Let quadrilateral $ABCD$ be inscribed in circle $O$ , where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at the point $H$ .

By the Pythagorean Theorem, the length of $OH$ is $\begin{align*} \sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2} \\ &= \sqrt{80000 - 10000} \\ &= \sqrt{70000} \\ &= 100\sqrt{7}. \end{align*}$

Note that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$ ; then we have that

$[AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$

Furthermore, $\begin{align*} h &= \sqrt{OD^2 - GD^2} \\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2} \\ &= \sqrt{80000 - \frac{x^2}{4}} \end{align*}$

Substituting this value of $h$ into the previous equation and evaluating for $x$ , we get: $\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}$ $\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}$ $60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)$ $40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}$ $400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}$ $(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}$ $7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)$ $7x^2 - 5600x + 1120000 = 320000 - x^2$ $8x^2 - 5600x + 800000 = 0$ $x^2 - 700x + 100000 = 0$

The roots of this quadratic are found by using the quadratic formula: $\begin{align*} x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} \\ &= 350 \pm \frac{300}{2} \\ &= 200, 500 \end{align*}$

If the length of $AD$ is $200$ , then quadrilateral $ABCD$ would be a square and thus, the radius of the circle would be $\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}$ Which is a contradiction. Therefore, our answer is $\boxed{500}.$

Solution 4 (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 $\cos\theta$ : $\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta$ $\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)$ $\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta$ It is useful to memorize the triple angle formulas ( $\cos 3\theta=4\cos^{3}\theta-3\cos\theta, \sin 3\theta=3\sin\theta-4\sin^{3}\theta$ ). 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 5 (Easier Trigonometry)

$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$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 desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$ , respectively. Also, let $\theta = \angle BOC$ . From the Law of Cosines on $\triangle BOC$ , we have $\cos \theta = \frac{3}{4}$ .

Now, since $\triangle BOC$ is isosceles with $\overline{OB} \cong \overline{OC}$ , we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$ . In addition, we know that $\overline{BC} \cong \overline{CD}$ as they are both equal to $200$ and $\overline{OB} \cong \overline{OC} \cong \overline{OD}$ as they are both radii of the same circle. 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 6 (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 $\triangle OBC$ to $\overline{OB}$ . Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$ , hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$ .

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

Solution 7 (Trigonometry)

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 8 (Area By Brahmagupta's Formula)

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 $\boxed{\textbf{(E)}\text{ 500}}$ .

Solution 9 (Similar Triangles)

$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations, L is used to write alpha= statement real RADIUS; pair A, B, C, D, E, F, O, L; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(A,D,O,B); F=extension(A,D,O,C); L=midpoint(C--D); O=(0,0); //Path Definitions path quad = A -- B -- C -- D -- cycle; //Initial Diagram draw(circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,NW); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,NE); label("$E$",E,SW); label("$F$",F,SE); label("$O$",O,SE); dot(O,linewidth(5)); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0)); draw(anglemark(C,O,B)); label("$\theta$",O,3N); draw(anglemark(E,F,O)); label("$\alpha$",F,3SW); draw(anglemark(D,F,C)); label("$\alpha$",F,3NE); draw(anglemark(F,C,D)); label("$\alpha$",C,3SSE); draw(anglemark(C,D,F)); label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW); [/asy]$ Label the points as shown, and let $\angle{EOF} = \theta$ . Since $\overline{OB} = \overline{OC}$ , and $\triangle{OFE} \sim \triangle{OCB}$ , we get that $\angle{EFO} = 90-\frac{\theta}{2}$ . We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity. From here, by vertical angles $\angle{CFD} = \alpha$ . Also, since $\triangle{OCB} \cong \triangle{ODC}$ , $\angle{OCD} = \alpha$ . This means that $\angle{CDF} = 180-2\alpha = \theta$ , which leads to $\triangle{OCB} \sim \triangle{DCF}$ . Since we know that $\overline{CD} = 200$ , $\overline{DF} = 200$ , and by similar reasoning $\overline{AE} = 200$ . Finally, again using similar triangles, we get that $\overline{CF} = 100\sqrt{2}$ , which means that $\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}$ . We can again apply similar triangles (or use Power of a Point) to get $\overline{EF} = 100$ , and finally $\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{\textbf{(E)}500}$ - ColtsFan10

Solution 10 (Parameshwara’s Formula for Circumradius)

Scale down by $100$ . We know that the semiperimeter of the quadrilateral is $\frac{(2 + 2 + 2 + x)}{2}$ where $x = \overline {AD}$ . Simplifying we get $\frac{6 + x}{2}$ . Now, the radius is $2\sqrt {2}$ , so $2\sqrt {2} = \frac{1}{4} \sqrt \frac {(4 + 2x)^{3}}{(\frac{2 + x}{2})^{3} (\frac {6 - x}{2})^{3}}$ .

Simplifying we get $x = 5$ . So the answer is $500$ .

Solution 11 (Complex Numbers)

We first scale down by a factor of $200\sqrt{2}$ . Let the vertices of the quadrilateral be $A$ , $B$ , $C$ , and $D$ , so that $AD$ is the length of the fourth side. We draw this in the complex plane so that $D$ corresponds to the complex number $1$ , and we let $C$ correspond to the complex number $z$ . Then, $A$ corresponds to $z^3$ and $B$ corresponds to $z^2$ . We are given that $\lvert z \rvert = 1$ and $\lvert z-1 \rvert = 1/\sqrt{2}$ , and we wish to find $\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert$ . Let $z=a+bi$ , where $a$ and $b$ are real numbers. Then, $a^2+b^2=1$ and $a^2-2a+1+b^2=1/2$ ; solving for $a$ and $b$ yields $a=3/4$ and $b=\sqrt{7}/4$ . Thus, $AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}$ . Scaling back up gives us a final answer of $\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{\textbf{(E)} 500}$ .

~ Leo.Euler

Video Solution by AoPS (Deven Ware)

https://www.youtube.com/watch?v=hpSyHZwsteM

Video Solution by Walt S.

https://www.youtube.com/watch?v=3iDqR9YNNkU

Video Solution (Ptolemy’s Theorem)

https://youtu.be/NsQbhYfGh1Q?t=5094

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/gCmQlaiEG5A

~IceMatrix

Video Solution by Punxsutawney Phil

https://www.youtube.com/watch?v=st6HIgDWgX4

@@ Line 424: / Line 424: @@
 ~ Leo.Euler
-==Remark (Morley's Trisector Theorem)==
-This problem is related to [http://www.cut-the-knot.org/triangle/Morley/Naraniengar.shtml M. T. Naraniengar's proof] of [https://en.wikipedia.org/wiki/Morley%27s_trisector_theorem Morley's Trisector Theorem]. This problem is taken from the figure of the Lemma of M. T. Naraniengar's proof, as shown below.
-[[File:NaraniengarLemma.gif|200px]]
-If four points <math>Y'</math>, <math>Z</math>, <math>Y</math>, <math>Z'</math> satisfy the conditions
-<math>\quad</math> <math>1.</math> <math>Y'Z = ZY = YZ'</math> and
-<math>\quad</math> <math>2.</math> <math>\angle YZY'</math> = <math>\angle Z'YZ</math> = <math>180^{\circ} - 2a > 60^{\circ}</math>
-then they lie on a circle.
-The Lemma is used to prove Morley's Trisector Theorem by constructing an equilateral triangle at <math>YZ</math> and extending <math>AY'</math> and <math>AZ'</math> as shown below.
-[[File:NaraniengarTheorem.gif|400px]]
-~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 ==Video Solution by AoPS (Deven Ware)==

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

Page

Toolbox

Search