2020 AMC 12B Problems/Problem 10
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Similar Triangles)
- 4 Solution 2 (Inscribed Angle Theorem and Pythagorean Theorem)
- 5 Solution 3 (Power of a Point)
- 6 Solution 4 (Trigonometry)
- 7 Solution 5 (Trigonometry)
- 8 Solution 6 (Coordinate Geometry)
- 9 Video Solution (HOW TO THINK CREATIVELY!!!)
- 10 Video Solution by TheBeautyOfMath
- 11 See Also
Problem
In unit square the inscribed circle intersects at and intersects at a point different from What is
Diagram
~MRENTHUSIASM
Solution 1 (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 using similar triangles ~QIDb602
Solution 2 (Inscribed Angle Theorem and Pythagorean Theorem)
Let be the midpoint of from which Note that by the Inscribed Angle Theorem.
We have the following diagram: Since and we get by the Pythagorean Theorem.
Let It follows that Applying the Pythagorean Theorem to right gives and applying the Pythagorean Theorem to right gives Equating the expressions for produces Finally, dividing both sides by and then rationalizing the denominator, we obtain
~MRENTHUSIASM
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
~Argonauts16
Solution 4 (Trigonometry)
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 5 (Trigonometry)
Take as the center and draw segment perpendicular to , , link . Then we have . So . Since , we have . As a result, Thus . Since , we have .
~FANYUCHEN20020715
Solution 6 (Coordinate Geometry)
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
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution by TheBeautyOfMath
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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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