Difference between revisions of "2020 AMC 12B Problems/Problem 10"
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Let circle <math>\omega</math> intersect <math>\overline{AB}</math> at point <math>N</math>. By Power of a Point, we have <math>AN^2=AP\cdot AM</math>. We know <math>AN=\frac{1}{2}</math> because <math>N</math> is the midpoint of <math>\overline{AB}</math>, and we can easily find <math>AM</math> by the Pythagorean Theorem, which gives us <math>AM=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}</math>. Our equation is now <math>\frac{1}{4}=AP\cdot \frac{\sqrt{5}}{2}</math>, or <math>AP=\frac{2}{4\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{2\cdot 5}</math>, thus our answer is | Let circle <math>\omega</math> intersect <math>\overline{AB}</math> at point <math>N</math>. By Power of a Point, we have <math>AN^2=AP\cdot AM</math>. We know <math>AN=\frac{1}{2}</math> because <math>N</math> is the midpoint of <math>\overline{AB}</math>, and we can easily find <math>AM</math> by the Pythagorean Theorem, which gives us <math>AM=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}</math>. Our equation is now <math>\frac{1}{4}=AP\cdot \frac{\sqrt{5}}{2}</math>, or <math>AP=\frac{2}{4\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{2\cdot 5}</math>, thus our answer is | ||
<math>\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.</math> | <math>\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.</math> | ||
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+ | ==Solution 3== | ||
+ | Take <math>O</math> as the center and draw segment <math>ON</math> perpendicular to <math>AM</math>. Then we have <math>OM\parallel AD</math>. So <math>\angle DAM=\angle OMA</math>. Since <math>AD=2AM=2OM=1</math>, we have <math>\cos\angle DAM=\cos\angle OMP=\frac{2}{\sqrt{5}}</math>. As a result, <math>NM=OM\cos\angle OMP=\frac{1}{2}\cdot \frac{2}{\sqrt{5}}=\frac{1}{\sqrt{5}}.</math> Thus <math>PM=2NM=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}</math>. Since <math>AM=\frac{\sqrt{5}}{2}</math>, we have <math>AP=AM-PM=\frac{\sqrt{5}}{10}</math>. Put <math>\boxed{B}</math>. | ||
+ | |||
+ | ~FANYUCHEN20020715 | ||
==See Also== | ==See Also== |
Revision as of 02:41, 8 February 2020
Contents
Problem
In unit square the inscribed circle intersects at and intersects at a point different from What is
Solution 1(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 2(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 3
Take as the center and draw segment perpendicular to . Then we have . So . Since , we have . As a result, Thus . Since , we have . Put .
~FANYUCHEN20020715
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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