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Difference between revisions of "2016 AMC 10A Problems/Problem 24"
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A quick aside -- this problem takes about 90 seconds to solve with an accurate compass and a good ruler. Draw the diagram, to scale, and measure the last side. It becomes very clear that the answer is <math>\boxed{\textbf{(E)}\text{ 500}}</math>.  
 
A quick aside -- this problem takes about 90 seconds to solve with an accurate compass and a good ruler. Draw the diagram, to scale, and measure the last side. It becomes very clear that the answer is <math>\boxed{\textbf{(E)}\text{ 500}}</math>.  
 +
-- Jonathan Ko
 +
 +
==Solution 11 (Very Cheap Nonsensical Guessing)==
 +
 +
(While this is very quick, it is not recommended in general.) Note that this is only a good idea if you have backup by the means of some other method, and can be used to check your answer by logic. On high-numbered AMC problems, answers are generally integers. Therefore, we can readily omit (B), (C), and (D). We have either (A) or (E). If (A) was correct, then the quadrilateral would be a square, which would be too predictable, and thus highly unlikely for a final AMC problem. You could also quickly check that if the quadrilateral was a square, then the center of the circle would be the center of the square. This would mean that the diameter of the circle would be the diagonal length of the square, namely <math>200\sqrt2</math>, and thus the radius would be <math>100\sqrt{2}</math>, contradicting the problem constraints. The only answer left is (E)
 +
 +
  
−
-- Jonathan Ko
 
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2016|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2016|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}
−
 
−
==Solution 11 (Very Cheap Nonsensical Guessing)--
 
−
 
−
(While this is very quick, it is not recommended in general.) Note that this is only a good idea if you have backup by the means of some other method, and can be used to check your answer by logic. On high-numbered AMC problems, answers are generally integers. Therefore, we can readily omit (B), (C), and (D). We have either (A) or (E). If (A) was correct, then the quadrilateral would be a square, which would be too predictable, and thus highly unlikely for a final AMC problem. You could also quickly check that if the quadrilateral was a square, then the center of the circle would be the center of the square. This would mean that the diameter of the circle would be the diagonal length of the square, namely <math>200\sqrt2</math>, and thus the radius would be <math>100\sqrt{2}</math>, contradicting the problem constraints. The only answer left is (E)
 

Revision as of 22:05, 9 July 2017

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 (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 (Basic 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 3 (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 $\cos\theta$: \[\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta\] \[\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-1+2\cos^{2}\theta)\] \[\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta\] 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 4 (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 5 (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.


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

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


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 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 $\triangle OBC$. 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 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 (Cheap Solution - For when you are running out of time.)

WLOG, let $AB=BC=CD=200$, and let ABCD be inscribed in a clrcle with radius $200\sqrt2$. We draw perpendiculars from $B$ and $C$ to $AD$, and label the intersections $E$ and $F$, respectively. We can see that $EF=200$ (because BCFE is a rectangle), and since $AD$ is clearly greater than 200, and and since $EF$, which is part of segment $AD$, is an integer, than we conclude that $AD$ is also an integer or of the form $200+2*AE$. There is no reason for $AE$ to be of the form $a\sqrt{b} - 100$ because it seems too arbitrary. The only other integer choice is $\boxed{\textbf{(E)}\text{ 500}}$.

Solution 10 (Measuring)

A quick aside -- this problem takes about 90 seconds to solve with an accurate compass and a good ruler. Draw the diagram, to scale, and measure the last side. It becomes very clear that the answer is $\boxed{\textbf{(E)}\text{ 500}}$. -- Jonathan Ko

Solution 11 (Very Cheap Nonsensical Guessing)

(While this is very quick, it is not recommended in general.) Note that this is only a good idea if you have backup by the means of some other method, and can be used to check your answer by logic. On high-numbered AMC problems, answers are generally integers. Therefore, we can readily omit (B), (C), and (D). We have either (A) or (E). If (A) was correct, then the quadrilateral would be a square, which would be too predictable, and thus highly unlikely for a final AMC problem. You could also quickly check that if the quadrilateral was a square, then the center of the circle would be the center of the square. This would mean that the diameter of the circle would be the diagonal length of the square, namely $200\sqrt2$, and thus the radius would be $100\sqrt{2}$, contradicting the problem constraints. The only answer left is (E)



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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