Difference between revisions of "2016 AMC 10A Problems/Problem 24"
Dandan2002 (talk | contribs) (→Solution 1) |
(New solution) |
||
Line 55: | Line 55: | ||
Construct quadrilateral <math>ABCD</math> 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 <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Apply law of cosines on triangle <math>OBC</math> with angle <math>BOC</math> as <math>\theta</math>. We get the following equation: <cmath>(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta</cmath> Substituting the values in, we get <cmath>(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta</cmath> Canceling out, we get <cmath>\cos\theta=\frac{3}{4}</cmath> | Construct quadrilateral <math>ABCD</math> 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 <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Apply law of cosines on triangle <math>OBC</math> with angle <math>BOC</math> as <math>\theta</math>. We get the following equation: <cmath>(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta</cmath> Substituting the values in, we get <cmath>(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta</cmath> Canceling out, we get <cmath>\cos\theta=\frac{3}{4}</cmath> | ||
To find the remaining side (<math>AD</math>), we simply have to apply law of cosines with angle <math>3\theta</math> on <math>OAD</math> since the other triangles <math>OAB</math>, <math>OBC</math>, and <math>OCD</math> are congruent. Now, to find <math>\cos 3\theta</math>, we can derive a formula that only uses <math>\cos\theta</math>: <cmath>\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- \sin 2\theta \cdot (2\sin\theta \cos\theta)</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-1+2\cos^{2}\theta)</cmath> <cmath>\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta</cmath> Plugging in <math>\cos\theta=\frac{3}{4}</math>, we get <math>\cos 3\theta= -\frac{9}{16}</math>. Now, applying law of cosines on triangle <math>OAD</math>, we get <cmath>(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}</cmath> <cmath>\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}</cmath> <cmath>AD=200 \cdot \frac{5}{2}=\boxed{500}</cmath> | To find the remaining side (<math>AD</math>), we simply have to apply law of cosines with angle <math>3\theta</math> on <math>OAD</math> since the other triangles <math>OAB</math>, <math>OBC</math>, and <math>OCD</math> are congruent. Now, to find <math>\cos 3\theta</math>, we can derive a formula that only uses <math>\cos\theta</math>: <cmath>\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- \sin 2\theta \cdot (2\sin\theta \cos\theta)</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-1+2\cos^{2}\theta)</cmath> <cmath>\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta</cmath> Plugging in <math>\cos\theta=\frac{3}{4}</math>, we get <math>\cos 3\theta= -\frac{9}{16}</math>. Now, applying law of cosines on triangle <math>OAD</math>, we get <cmath>(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}</cmath> <cmath>\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}</cmath> <cmath>AD=200 \cdot \frac{5}{2}=\boxed{500}</cmath> | ||
+ | |||
+ | ==Solution 3 (Easier trig)== | ||
+ | <Diagram Needed> | ||
+ | Construct quadrilateral <math>ABCD</math> on the circle <math>O</math> with <math>AD</math> being the missing side. Then, drop perpendiculars from <math>A</math> and <math>D</math> to (extended) line <math>BC</math>, and let these points be <math>E</math> and <math>F</math>, respectively. Also, let <math>\theta = \angle BOC</math>. From Law of Cosines on <math>\triangle BOC</math>, we have <math>\cos \theta = \frac{3}{4}</math>. | ||
+ | Now, since <math>\triangle OBC</math> is isosceles with <math>OB = OC</math>, we have that <math>\angle BCO = \angle CBO = 90 - \frac{\theta}{2}</math>. By SSS congruence, we have that <math>\triangle OBC \cong \triangle OCD</math>, so we have that <math>\angle OCD = \angle CBO = 90 - \frac{\theta}{2}</math>, so <math>\angle DFC = \theta</math>. | ||
+ | Thus, we have <math>\frac{FC}{DC} = \cos \theta = \frac{3}{4}</math>, so <math>FC = 150</math>. Similarly, <math>BE = 150</math>, and <math>AD = 150 + 200 + 150 = \boxed{500}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}} | {{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:32, 3 February 2016
Contents
Problem
A quadrilateral is inscribed in a circle of radius . Three of the sides of this quadrilateral have length . What is the length of the fourth side?
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 for now, then multiply it back at the end of our solution.
Construct quadrilateral on the circle with 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 to and . Let the intersection of and be point . Notice that and are perpendicular because is a kite.
We set lengths equal to . By the Pythagorean Theorem,
We solve for :
.
By Ptolemy's Theorem,
Substituting values,
Finally, we multiply back the that we divided by at the beginning of the problem to get .
Solution 2 (Trigonometry Bash)
<Diagram Needed>
Construct quadrilateral 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 to and . Apply law of cosines on triangle with angle as . We get the following equation: Substituting the values in, we get Canceling out, we get To find the remaining side (), we simply have to apply law of cosines with angle on since the other triangles , , and are congruent. Now, to find , we can derive a formula that only uses : Plugging in , we get . Now, applying law of cosines on triangle , we get
Solution 3 (Easier trig)
<Diagram Needed>
Construct quadrilateral on the circle with being the missing side. Then, drop perpendiculars from and to (extended) line , and let these points be and , respectively. Also, let . From Law of Cosines on , we have . Now, since is isosceles with , we have that . By SSS congruence, we have that , so we have that , so . Thus, we have , so . Similarly, , and .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.