2020 AMC 12B Problems/Problem 12
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Pythagorean Theorem)
- 4 Solution 2 (Power of a Point)
- 5 Solution 3 (Law of Cosines)
- 6 Solution 4 (Reflections)
- 7 Solution 5 (Basically Solution 2 With Motivation)
- 8 Solution 6 (Double Power of a Point)
- 9 Solution 7 (Cheating)
- 10 Video Solution by OmegaLearn
- 11 Video Solution by On The Spot STEM
- 12 Video Solution by TheBeautyOfMath
- 13 See Also
Problem
Let be a diameter in a circle of radius Let be a chord in the circle that intersects at a point such that and What is
Diagram
~Shihan ~MRENTHUSIASM
Solution 1 (Pythagorean Theorem)
Let be the center of the circle, and be the midpoint of . Let and . This implies that . Since , we now want to find . Since is a right angle, by Pythagorean theorem . Thus, our answer is .
~JHawk0224
Solution 2 (Power of a Point)
Let be the center of the circle, and be the midpoint of . Draw triangle , and median . Because , is isosceles, so is also an altitude of . , and because angle is degrees and triangle is right, . Because triangle is right, . Thus, .
We are looking for + which is also .
Because , .
By Power of a Point, , so .
Finally, .
Solution 3 (Law of Cosines)
Let be the center of the circle. Notice how , where is the radius of the circle. By applying the law of cosines on triangle ,
Similarly, by applying the law of cosines on triangle ,
By subtracting these two equations, we get We can rearrange it to get
Because both and are both positive, we can safely divide both sides by to obtain . Because ,
Through power of a point, we can find out that , so
~Math_Wiz_3.14 (legibility changes by eagleye)
Solution 4 (Reflections)
Let be the center of the circle. By reflecting across the line to produce , we have that . Since , . Since , by the Pythagorean Theorem, our desired solution is just . Looking next to circle arcs, we know that , so . Since , and , . Thus, . Since , by the Pythagorean Theorem, the desired .
~sofas103
Solution 5 (Basically Solution 2 With Motivation)
Basically, by PoP, you have that Therefore, as basically, once you find the problem is done. Now, this is an IMPORTANT concept: If you have a circle which you know the radius of and you want to find the length of a chord of that circle, drop an altitude from the center of the circle to the chord to find distance between the center of the circle and the chord.
In this case, let be the midpoint of chord Notice that now we can use our angle, since is a triangle so that and However, we have that so that Now, notice that so that and Therefore,
This may not be the “shortest solution”, but in my opinion is very well motivated and doesn’t require much creativity. [Not requiring much creativity, it also saves more time than you’d think. ;)]
~ Professor-Mom
Solution 6 (Double Power of a Point)
For ease of notation, let and . Extend to point until is perpendicular to . It's given that , so, by vertical angles, we have .
Since is a right triangle, we have and . Hence,
By Power of a Point, we have
Isolating the variables after expanding gives
Using Power of a Point again, we have
To get , we can perform the operation . Plugging these values in,
-Benedict T (countmath1)
Solution 7 (Cheating)
Perhaps not reliable in general, but very useful as a last resort. The choice of the radius is strange, and is probably motivated by a nice answer in the end, so we only consider integer options. Notice that a 5 also appears in the condition , therefore it will likely be present in the answer as well; the only integer containing a factor of 5 amongst the choices is 100, thus the answer is
~Maths357
Video Solution by OmegaLearn
https://youtu.be/LpOegT0fKy8?t=228
~ pi_is_3.14
Video Solution by On The Spot STEM
https://www.youtube.com/watch?v=h-hhRa93lK4
Video Solution by TheBeautyOfMath
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.