Difference between revisions of "2020 AMC 12B Problems/Problem 10"
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==Problem== | ==Problem== | ||
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In unit square <math>ABCD,</math> the inscribed circle <math>\omega</math> intersects <math>\overline{CD}</math> at <math>M,</math> and <math>\overline{AM}</math> intersects <math>\omega</math> at a point <math>P</math> different from <math>M.</math> What is <math>AP?</math> | In unit square <math>ABCD,</math> the inscribed circle <math>\omega</math> intersects <math>\overline{CD}</math> at <math>M,</math> and <math>\overline{AM}</math> intersects <math>\omega</math> at a point <math>P</math> different from <math>M.</math> What is <math>AP?</math> | ||
<math>\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}</math> | <math>\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}</math> | ||
+ | |||
+ | ==Diagram== | ||
+ | [[File:2020 AMC 12B Problem 12 Diagram.png|center]] | ||
+ | |||
+ | ~MRENTHUSIASM (by Geometry Expressions) | ||
==Solution 1 (Angle Chasing/Trig)== | ==Solution 1 (Angle Chasing/Trig)== | ||
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~awesome1st | ~awesome1st | ||
− | ==Solution 2(Coordinate Bash)== | + | |
+ | ==Solution 2 (Coordinate Bash)== | ||
Place circle <math>\omega</math> in the Cartesian plane such that the center lies on the origin. Then we can easily find the equation for <math>\omega</math> as <math>x^2+y^2=\frac{1}{4}</math>, because it is not translated and the radius is <math>\frac{1}{2}</math>. | Place circle <math>\omega</math> in the Cartesian plane such that the center lies on the origin. Then we can easily find the equation for <math>\omega</math> as <math>x^2+y^2=\frac{1}{4}</math>, because it is not translated and the radius is <math>\frac{1}{2}</math>. | ||
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~Argonauts16 | ~Argonauts16 | ||
− | ==Solution 3(Power of a Point)== | + | ==Solution 3 (Power of a Point)== |
− | |||
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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~QIDb602 | ~QIDb602 | ||
− | + | ==Solution 6 (Intersecting Chords)== | |
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− | |||
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− | ==Solution 6 | ||
Label the midpoint of <math>AD</math> as <math>N</math>, and let <math>Q</math> be the intersection of <math>ON</math> and <math>AM</math> | Label the midpoint of <math>AD</math> as <math>N</math>, and let <math>Q</math> be the intersection of <math>ON</math> and <math>AM</math> | ||
Then <math>MQ=AQ=\frac{1}{2}AM = \frac{\sqrt{5}}{4}</math> and <math>NQ=\frac{1}{4}</math> | Then <math>MQ=AQ=\frac{1}{2}AM = \frac{\sqrt{5}}{4}</math> and <math>NQ=\frac{1}{4}</math> | ||
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~ERMSCoach | ~ERMSCoach | ||
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+ | ==Solution 7 (Inscribed Angle Theorem and Pythagorean Theorem)== | ||
+ | Let <math>N</math> be the midpoint of <math>\overline{AB},</math> from which <math>\angle ANM=90^\circ.</math> Note that <math>\angle NPM=90^\circ</math> by the Inscribed Angle Theorem. | ||
+ | |||
+ | We have the following diagram: | ||
+ | |||
+ | [[File:2020 AMC 12B Problem 10 Solution.png|center]] | ||
+ | |||
+ | Since <math>AN=\frac12</math> and <math>NM=1,</math> we get <math>AM=\frac{\sqrt5}{2}</math> by the Pythagorean Theorem. | ||
+ | |||
+ | Let <math>AP=x.</math> It follows that <math>PM=\frac{\sqrt5}{2}-x.</math> Applying the Pythagorean Theorem to right <math>\triangle ANP</math> gives <math>NP^2=\left(\frac12\right)^2-x^2,</math> and applying the Pythagorean Theorem to right <math>\triangle MNP</math> gives <math>NP^2=1^2-\left(\frac{\sqrt5}{2}-x\right)^2.</math> Equating the expressions for <math>NP^2</math> produces | ||
+ | <cmath>\begin{align*} | ||
+ | \left(\frac12\right)^2-x^2&=1^2-\left(\frac{\sqrt5}{2}-x\right)^2 \\ | ||
+ | \frac14-x^2&=1-\frac54+\sqrt5x-x^2 \\ | ||
+ | \frac12&=\sqrt5x. | ||
+ | \end{align*}</cmath> | ||
+ | Finally, dividing both sides by <math>\sqrt5</math> then rationalizing the denominator, we obtain <math>x=\frac{1}{2\sqrt5}=\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/6ujfjGLzVoE | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Latest revision as of 10:30, 22 April 2021
Contents
Problem
In unit square the inscribed circle intersects at and intersects at a point different from What is
Diagram
~MRENTHUSIASM (by Geometry Expressions)
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.
Using the Pythagorean theorem, . Now we just need to find using similar triangles. ~QIDb602
Solution 6 (Intersecting Chords)
Label the midpoint of as , and let be the intersection of and Then and
Then and
~ERMSCoach
Solution 7 (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 then rationalizing the denominator, we obtain
~MRENTHUSIASM
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.