2020 AMC 12B Problems/Problem 10
Contents
Problem
In unit square the inscribed circle intersects at and intersects at a point different from What is
Solution 1 (Angle Chasing/Trig)
Let be the center of the circle and the point of tangency between and be represented by . We know that . Consider the right triangle . Let .
Since is tangent to at , . Now, consider . This triangle is iscoceles because and are both radii of . Therefore, .
We can now use Law of Cosines on to find the length of and subtract it from the length of to find . Since and , the double angle formula tells us that . We have By Pythagorean theorem, we find that
~awesome1st
Solution 2(Coordinate Bash)
Place circle in the Cartesian plane such that the center lies on the origin. Then we can easily find the equation for as , because it is not translated and the radius is .
We have and . The slope of the line passing through these two points is , and the -intercept is simply . This gives us the line passing through both points as .
We substitute this into the equation for the circle to get , or . Simplifying gives . The roots of this quadratic are and , but if we get point , so we only want .
We plug this back into the linear equation to find , and so . Finally, we use distance formula on and to get .
~Argonauts16
Solution 3(Power of a Point)
Let circle intersect at point . By Power of a Point, we have . We know because is the midpoint of , and we can easily find by the Pythagorean Theorem, which gives us . Our equation is now , or , thus our answer is
Solution 4
Take as the center and draw segment perpendicular to , , link . Then we have . So . Since , we have . As a result, Thus . Since , we have . Put .
~FANYUCHEN20020715
Video Solution
~IceMatrix
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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