2016 AMC 10A Problems/Problem 24
- The following problem is from both the 2016 AMC 10A #24 and 2016 AMC 12A #21, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Algebra)
- 4 Solution 3 (HARD Algebra)
- 5 Solution 4 (Trigonometry Bash)
- 6 Solution 5 (Easier Trigonometry)
- 7 Solution 6 (Ptolemy's Theorem)
- 8 Solution 7 (Trigonometry)
- 9 Solution 8 (Area By Brahmagupta's Formula)
- 10 Solution 9 (Similar Triangles)
- 11 Solution 10 (Complex Numbers)
- 12 Solution 11 ( Different Trigbash + Ptolmey’s)
- 13 Remark (Morley's Trisector Theorem)
- 14 Video Solution by AoPS (Deven Ware)
- 15 Video Solution by Walt S.
- 16 Video Solution (Ptolemy’s Theorem)
- 17 Video Solution by TheBeautyofMath
- 18 Video Solution by Punxsutawney Phil
- 19 Video Solution by OmegaLearn
- 20 See Also
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
Let intersect at and at
From there, , thus:
because they are both radii of . Since , we have that . Similarly, .
and , so
Solution 2 (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 (Solution 1.1 begins from here). 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 3 (HARD Algebra)
Let quadrilateral be inscribed in circle , where is the side of unknown length. Draw the radii from center to all four vertices of the quadrilateral, and draw the altitude of such that it passes through side at the point and meets side at the point .
By the Pythagorean Theorem, the length of is
Note that Let the length of be and the length of be ; then we have that
Furthermore,
Substituting this value of into the previous equation and evaluating for , we get:
The roots of this quadratic are found by using the quadratic formula:
If the length of is , then quadrilateral would be a square and thus, the radius of the circle would be Which is a contradiction. Therefore, our answer is
Solution 4 (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 : It is useful to memorize the triple angle formulas (). Plugging in , we get . Now, applying law of cosines on triangle , we get
Solution 5 (Easier Trigonometry)
Construct quadrilateral on the circle with being the desired side. Then, drop perpendiculars from and to the extended line of and let these points be and , respectively. Also, let . From the Law of Cosines on , we have .
Now, since is isosceles with , we have that . In addition, we know that as they are both equal to and as they are both radii of the same circle. By SSS Congruence, we have that , so we have that , so .
Thus, we have , so . Similarly, , and .
Solution 6 (Ptolemy's Theorem)
Let . Let be the center of the circle. Then is twice the altitude of to . Since is isosceles we can compute its area to be , hence .
Now by Ptolemy's Theorem we have This gives us:
Solution 7 (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 8 (Area By Brahmagupta's Formula)
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 .
Solution 9 (Similar Triangles)
Label the points as shown, and let . Since , and , we get that . We assign to for simplicity. From here, by vertical angles . Also, since , . This means that , which leads to . Since we know that , , and by similar reasoning . Finally, again using similar triangles, we get that , which means that . We can again apply similar triangles (or use Power of a Point) to get , and finally - ColtsFan10
Solution 10 (Complex Numbers)
We first scale down by a factor of . Let the vertices of the quadrilateral be , , , and , so that is the length of the fourth side. We draw this in the complex plane so that corresponds to the complex number , and we let correspond to the complex number . Then, corresponds to and corresponds to . We are given that and , and we wish to find . Let , where and are real numbers. Then, and ; solving for and yields and . Thus, . Scaling back up gives us a final answer of .
~ Leo.Euler
Solution 11 ( Different Trigbash + Ptolmey’s)
Let angle be . This way will be . Now we can begin the trigbash. As the circumradius of triangle is , we can use the formula and and plug in all the values we got to get . This boils down to . This expression can further be simplified by the trig identity . This leads to the final simplified form . Solving this expression gives us . However, as we want , we use the even cooler trig identity , and substitute the given values to get that , and therefore BD is .
After we get BD we can move on to the next step with Ptolmey’s. As is a cyclic quadrilateral, we can use ptolmey’s theorem (with ) to get . Finally solve this equation to get that , or
-dragoon
Remark (Morley's Trisector Theorem)
This problem is related to M. T. Naraniengar's proof of Morley's Trisector Theorem. This problem is taken from the figure of the Lemma of M. T. Naraniengar's proof, as shown below.
If four points , , , satisfy the conditions
and
= =
then they lie on a circle.
The Lemma is used to prove Morley's Trisector Theorem by constructing an equilateral triangle at and extending and as shown below.
Video Solution by AoPS (Deven Ware)
https://www.youtube.com/watch?v=hpSyHZwsteM
Video Solution by Walt S.
https://www.youtube.com/watch?v=3iDqR9YNNkU
Video Solution (Ptolemy’s Theorem)
https://youtu.be/NsQbhYfGh1Q?t=5094
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=st6HIgDWgX4
Video Solution by OmegaLearn
https://youtu.be/NsQbhYfGh1Q?t=5094
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.