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Difference between revisions of "2020 AMC 12B Problems/Problem 10"

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~Argonauts16
 
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==Video Solution==
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==Video Solution (HOW TO THINK CREATIVELY!!!)==
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https://youtu.be/gd6VvCKMdu4
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~Education, the Study of Everything
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==Video Solution by TheBeautyOfMath==
 
https://youtu.be/6ujfjGLzVoE
 
https://youtu.be/6ujfjGLzVoE
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~IceMatrix
 
  
 
==See Also==
 
==See Also==

Latest revision as of 15:14, 11 September 2023

Problem

In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$

$\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}$

Diagram

[asy] /* Made by MRENTHUSIASM */ size(180); pair A, B, C, D, M, O, P; O = origin; A = (-1/2,-1/2); B = (-1/2,1/2); C = (1/2,1/2); D = (1/2,-1/2); M = midpoint(C--D); path p; p = Circle(O,1/2); P = intersectionpoints(A--M,p)[0]; dot("$\omega$",O,1.5*N,linewidth(4)); dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*NE,linewidth(4)); dot("$D$",D,1.5*SE,linewidth(4)); dot("$M$",M,1.5*E,linewidth(4)); dot("$P$",P,1.5*N,linewidth(4)); draw(A--B--C--D--cycle^^A--M^^p); [/asy] ~MRENTHUSIASM

Solution 1 (Similar Triangles)

Call the midpoint of $\overline{AB}$ point $N.$ Draw in $\overline{NM}$ and $\overline{NP}.$ Note that $\angle{NPM}=90^{\circ}$ due to Thales's Theorem. [asy] /* Made by QIDb602; edited by MRENTHUSIASM */ size(180); pair A, B, C, D, M, N, O, P; O = origin; A = (-1/2,-1/2); B = (-1/2,1/2); C = (1/2,1/2); D = (1/2,-1/2); M = midpoint(C--D); N = midpoint(A--B); path p; p = Circle(O,1/2); P = intersectionpoints(A--M,p)[0]; fill(N--A--M--cycle,yellow); dot("$\omega$",O,1.5*(0,1),linewidth(4)); dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*NE,linewidth(4)); dot("$D$",D,1.5*SE,linewidth(4)); dot("$M$",M,1.5*E,linewidth(4)); dot("$N$",N,1.5*W,linewidth(4)); dot("$P$",P,1.5*dir(60),linewidth(4)); markscalefactor=0.00625; draw(rightanglemark(A,N,M),red); draw(rightanglemark(N,P,A),red); draw(A--B--C--D--cycle^^A--M^^P--N--M^^p); [/asy] Using the Pythagorean theorem, $AM=\frac{\sqrt{5}}{2}.$ Now we just need to find $AP$ using similar triangles $\triangle APN\sim\triangle ANM:$ \begin{align*} \frac{AP}{AN}&=\frac{AN}{AM} \\ \frac{AP}{1/2}&=\frac{1/2}{\sqrt5/2} \\ AP&=\boxed{\textbf{(B) } \frac{\sqrt5}{10}}. \end{align*} ~QIDb602

Solution 2 (Inscribed Angle Theorem and Pythagorean Theorem)

Let $N$ be the midpoint of $\overline{AB},$ from which $\angle ANM=90^\circ.$ Note that $\angle NPM=90^\circ$ by the Inscribed Angle Theorem.

We have the following diagram: [asy] /* Made by MRENTHUSIASM */ size(180); pair A, B, C, D, M, N, O, P; O = origin; A = (-1/2,-1/2); B = (-1/2,1/2); C = (1/2,1/2); D = (1/2,-1/2); M = midpoint(C--D); N = midpoint(A--B); path p; p = Circle(O,1/2); P = intersectionpoints(A--M,p)[0]; fill(N--P--A--cycle,yellow); fill(N--P--M--cycle,green); dot("$\omega$",O,1.5*(0,1),linewidth(4)); dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*NE,linewidth(4)); dot("$D$",D,1.5*SE,linewidth(4)); dot("$M$",M,1.5*E,linewidth(4)); dot("$N$",N,1.5*W,linewidth(4)); dot("$P$",P,1.5*dir(60),linewidth(4)); markscalefactor=0.00625; draw(rightanglemark(A,N,M),red); draw(rightanglemark(N,P,A),red); draw(A--B--C--D--cycle^^A--M^^P--N--M^^p); [/asy] Since $AN=\frac12$ and $NM=1,$ we get $AM=\frac{\sqrt5}{2}$ by the Pythagorean Theorem.

Let $AP=x.$ It follows that $PM=\frac{\sqrt5}{2}-x.$ Applying the Pythagorean Theorem to right $\triangle ANP$ gives $NP^2=\left(\frac12\right)^2-x^2,$ and applying the Pythagorean Theorem to right $\triangle MNP$ gives $NP^2=1^2-\left(\frac{\sqrt5}{2}-x\right)^2.$ Equating the expressions for $NP^2$ produces \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*} Finally, dividing both sides by $\sqrt5$ and then rationalizing the denominator, we obtain $x=\frac{1}{2\sqrt5}=\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.$

~MRENTHUSIASM

Solution 3 (Power of a Point)

Let circle $\omega$ intersect $\overline{AB}$ at point $N$. By Power of a Point, we have $AN^2=AP\cdot AM$. We know $AN=\frac{1}{2}$ because $N$ is the midpoint of $\overline{AB}$, and we can easily find $AM$ by the Pythagorean Theorem, which gives us $AM=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}$. Our equation is now $\frac{1}{4}=AP\cdot \frac{\sqrt{5}}{2}$, or $AP=\frac{2}{4\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{2\cdot 5}$, thus our answer is $\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.$

~Argonauts16

Solution 4 (Trigonometry)

Let $O$ be the center of the circle and the point of tangency between $\omega$ and $\overline{AD}$ be represented by $K$. We know that $\overline{AK} = \overline{KD} = \overline{DM} = \frac{1}{2}$. Consider the right triangle $\bigtriangleup ADM$. Let $\measuredangle AMD = \theta$.

Since $\omega$ is tangent to $\overline{DC}$ at $M$, $\measuredangle PMO = 90 - \theta$. Now, consider $\bigtriangleup POM$. This triangle is iscoceles because $\overline{PO}$ and $\overline{OM}$ are both radii of $\omega$. Therefore, $\measuredangle POM = 180 - 2(90 - \theta) = 2\theta$.

We can now use Law of Cosines on $\angle{POM}$ to find the length of ${PM}$ and subtract it from the length of ${AM}$ to find ${AP}$. Since $\cos{\theta} = \frac{1}{\sqrt{5}}$ and $\sin{\theta} = \frac{2}{\sqrt{5}}$, the double angle formula tells us that $\cos{2\theta} = -\frac{3}{5}$. We have \[PM^2 = \frac{1}{2} - \frac{1}{2}\cos{2\theta} \implies PM = \frac{2\sqrt{5}}{5}.\] By Pythagorean theorem, we find that $AM = \frac{\sqrt{5}}{2} \implies AP=\boxed{\textbf{(B) } \frac{\sqrt5}{10}}$.

~awesome1st

Solution 5 (Trigonometry)

Take $O$ as the center and draw segment $ON$ perpendicular to $AM$, $ON\cap AM=N$, link $OM$. Then we have $OM\parallel AD$. So $\angle DAM=\angle OMA$. Since $AD=2AM=2OM=1$, we have $\cos\angle DAM=\cos\angle OMP=\frac{2}{\sqrt{5}}$. As a result, $NM=OM\cos\angle OMP=\frac{1}{2}\cdot \frac{2}{\sqrt{5}}=\frac{1}{\sqrt{5}}.$ Thus $PM=2NM=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$. Since $AM=\frac{\sqrt{5}}{2}$, we have $AP=AM-PM=\boxed{\textbf{(B) } \frac{\sqrt5}{10}}$.

~FANYUCHEN20020715

Solution 6 (Coordinate Geometry)

Place circle $\omega$ in the Cartesian plane such that the center lies on the origin. Then we can easily find the equation for $\omega$ as $x^2+y^2=\frac{1}{4}$, because it is not translated and the radius is $\frac{1}{2}$.

We have $A=\left(-\frac{1}{2}, \frac{1}{2}\right)$ and $M=\left(0, -\frac{1}{2}\right)$. The slope of the line passing through these two points is $\frac{\frac{1}{2}+\frac{1}{2}}{-\frac{1}{2}-0}=\frac{1}{-\frac{1}{2}}=-2$, and the $y$-intercept is simply $M$. This gives us the line passing through both points as $y=-2x-\frac{1}{2}$.

We substitute this into the equation for the circle to get $x^2+\left(-2x-\frac{1}{2}\right)^2=\frac{1}{4}$, or $x^2+4x^2+2x+\frac{1}{4}=\frac{1}{4}$. Simplifying gives $x(5x+2)=0$. The roots of this quadratic are $x=0$ and $x=-\frac{2}{5}$, but if $x=0$ we get point $M$, so we only want $x=-\frac{2}{5}$.

We plug this back into the linear equation to find $y=\frac{3}{10}$, and so $P=\left(-\frac{2}{5}, \frac{3}{10}\right)$. Finally, we use distance formula on $A$ and $P$ to get $AP=\sqrt{\left(-\frac{5}{10}+\frac{4}{10}\right)^2+\left(\frac{5}{10}-\frac{3}{10}\right)^2}=\sqrt{\frac{1}{100}+\frac{4}{100}}=\boxed{\textbf{(B) } \frac{\sqrt5}{10}}$.

~Argonauts16

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/gd6VvCKMdu4

~Education, the Study of Everything




Video Solution by TheBeautyOfMath

https://youtu.be/6ujfjGLzVoE

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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