Difference between revisions of "2016 AMC 10A Problems/Problem 24"
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Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> 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 <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Let the intersection of <math>BD</math> and <math>OC</math> be point <math>E</math>. Notice that <math>BD</math> and <math>OC</math> are perpendicular because <math>BCDO</math> is a kite. | Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> 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 <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Let the intersection of <math>BD</math> and <math>OC</math> be point <math>E</math>. Notice that <math>BD</math> and <math>OC</math> are perpendicular because <math>BCDO</math> is a kite. | ||
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We set lengths <math>BE=ED</math> equal to <math>x</math>. By the Pythagorean Theorem, | We set lengths <math>BE=ED</math> equal to <math>x</math>. By the Pythagorean Theorem, | ||
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<cmath>\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}</cmath> | <cmath>\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}</cmath> | ||
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We solve for <math>x</math>: | We solve for <math>x</math>: | ||
<cmath>1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2</cmath> | <cmath>1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2</cmath> | ||
<cmath>2\sqrt{(1-x^2)(2-x^2)}=2x^2-1</cmath> | <cmath>2\sqrt{(1-x^2)(2-x^2)}=2x^2-1</cmath> | ||
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<cmath>4(1-x^2)(2-x^2)=(2x^2-1)^2</cmath> | <cmath>4(1-x^2)(2-x^2)=(2x^2-1)^2</cmath> | ||
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<cmath>8-12x^2+4x^4=4x^4-4x^2+1</cmath> | <cmath>8-12x^2+4x^4=4x^4-4x^2+1</cmath> | ||
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<cmath>8x^2=7</cmath> | <cmath>8x^2=7</cmath> | ||
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<cmath>x=\frac{\sqrt{14}}{4}</cmath> | <cmath>x=\frac{\sqrt{14}}{4}</cmath> | ||
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By Ptolemy's Theorem, | By Ptolemy's Theorem, | ||
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<cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2</cmath> | <cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2</cmath> | ||
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Substituting values, | Substituting values, | ||
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<cmath>1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2</cmath> | <cmath>1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2</cmath> | ||
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<cmath>1+AD=\frac{7}{2}</cmath> | <cmath>1+AD=\frac{7}{2}</cmath> | ||
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<cmath>AD=\frac{5}{2}</cmath> | <cmath>AD=\frac{5}{2}</cmath> | ||
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Finally, we multiply back the <math>200</math> that we divided by at the beginning of the problem to get <math>AD=\boxed{500 (E)}</math>. | Finally, we multiply back the <math>200</math> that we divided by at the beginning of the problem to get <math>AD=\boxed{500 (E)}</math>. |
Revision as of 13:38, 4 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
Solution 1 (Algebra)
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 top 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)
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 top of the circle). Now, draw the radii from center to and . Apply law of cosines on ; let . We get the following equation: Substituting the values in, we get Canceling out, we get Because , , and are congruent, . To find the remaining side (), we simply have to apply the law of cosines to . 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)
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 |
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