2020 AMC 12B Problems/Problem 10
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
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 .
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
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 .
Solution 5 (Similar Triangles)
Call the midpoint of point . Draw in and . Note that due to Thales's Theorem.
Using the Pythagorean theorem, . Now we just need to find the length by setting up proportions on similar triangles. ~QIDb602
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