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2016 AMC 10A Problems/Problem 24

Revision as of 21:32, 3 February 2016 by VinVinB (talk | contribs) (New solution)

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

<Diagram and Reformatting Needed>


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.


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 bottom 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 $AD=\boxed{500}$.

Solution 2 (Trigonometry Bash)

<Diagram Needed>

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 bottom of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply law of cosines on triangle $OBC$ with angle $BOC$ as $\theta$. 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 \[\cos\theta=\frac{3}{4}\] To find the remaining side ($AD$), we simply have to apply law of cosines with angle $3\theta$ on $OAD$ since the other triangles $OAB$, $OBC$, and $OCD$ are congruent. 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)

<Diagram Needed>

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 $\triangle OBC$ 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 $\angle OCD = \angle CBO = 90 - \frac{\theta}{2}$, so $\angle DFC = \theta$. Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \boxed{500}$.

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

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