Difference between revisions of "2020 AMC 12B Problems/Problem 10"
Argonauts16 (talk | contribs) (→Solution 3(Power of a Point)) |
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~FANYUCHEN20020715 | ~FANYUCHEN20020715 | ||
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+ | ==Solution 5 (Similar Triangles)== | ||
+ | Call the midpoint of <math>\overline{AB}</math> point <math>N</math>. Draw in <math>\overline{NM}</math> and <math>\overline{NP}</math>. Note that <math>\angle{NPM}=90^{\circ}</math> due to Thales's Theorem. | ||
+ | |||
+ | <asy> | ||
+ | draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); | ||
+ | draw(circle((0.5,0.5),0.5)); | ||
+ | draw((0,0)--(0,0.5)--(1,0.5)--cycle); | ||
+ | label("A",(0,0),SW); | ||
+ | label("B",(0,1),NW); | ||
+ | label("C",(1,1),NE); | ||
+ | label("D",(1,0),SE); | ||
+ | label("M",(1,0.5),E); | ||
+ | label("P",(0.2,0.1),S); | ||
+ | label("N",(0,0.5),W); | ||
+ | draw((0,0.5)--(0.2,0.1)); | ||
+ | markscalefactor=0.007; | ||
+ | draw(rightanglemark((0,0.5),(0.2,0.1),(1,0.5))); | ||
+ | </asy> | ||
+ | Now we just need to find the length by setting up proportions on similar triangles. | ||
+ | <cmath>\triangle APN\sim\triangle ANM\Rightarrow\frac{AP}{AN}=\frac{AN}{AM}\Rightarrow\frac{AP}{\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}}</cmath> | ||
+ | <cmath>AP=\boxed{\textbf{(B) }\frac{\sqrt{5}}{10}}</cmath> | ||
+ | ~QIDb602 | ||
==Video Solution== | ==Video Solution== |
Revision as of 15:04, 24 February 2020
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
~Argonauts16
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
Solution 5 (Similar Triangles)
Call the midpoint of point . Draw in and . Note that due to Thales's Theorem.
Now we just need to find the length by setting up proportions on similar triangles. ~QIDb602
Video Solution
~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.