2016 AMC 10A Problems/Problem 24
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 .
Solution 4 (Just Geometry)
Label AD intercept OB at E and OC at F.
From there, , thus:
because they are both radii of . Since , we have that . Similarly, .
and , so
Solution 5 (Ptolemy's Theorem)
Let . Let be the center of the circle. Then is twice the altitude of . Since is isosceles we can compute its area to be , hence .
Now by Ptolemy's Theorem we have This gives us:
Solution 6 (Trigonometry)
Since all three sides equal , 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 by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is . Similarly, the cosine is . Since there are three sides, and since ,we seek to find . First, and by Pythagorean.
Solution 7 (Area)
For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be , where and is the missing side length. Let . If and are the midpoints of and , respectively, the height of the trapezoid is . By the pythagorean theorem, and . Thus the height of the trapezoid is , so the area is . By Brahmagupta's formula, the area is . Setting these two equal, we get . Dividing both sides by and then squaring, we get . Expanding the right hand side and canceling the terms gives us . Rearranging and dividing by two, we get . Squaring both sides, we get . Rearranging, we get . Dividing by 4 we get . Factoring we get, , and since cannot be negative, we get . Since , . Scaling up by 100, we get .
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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