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2025 AMC 12A Problems/Problem 10: Difference between revisions

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==See Also==
==See Also==
{{AMC10 box|year=2025|ab=A|before=First Problem|num-a=2}}
{{AMC12 box|year=2025|ab=A|num-b=9|num-a=11}}
{{AMC12 box|year=2025|ab=A|before=First Problem|num-a=2}}
* [[AMC 10]]
* [[AMC 10 Problems and Solutions]]
* [[Mathematics competitions]]
* [[Mathematics competition resources]]
{{MAA Notice}}
{{MAA Notice}}

Revision as of 16:15, 6 November 2025

Problem

In the figure shown below, major arc $\widehat{AD}$ and minor arc $\widehat{BC}$ have the same center, $O$. Also, $A$ lies between $O$ and $B$, and $D$ lies between $O$ and $C$. Major arc $\widehat{AD}$, minor arc $\widehat{BC}$, and each of the two segments $\overline{AB}$ and $\overline{CD}$ have length $2\pi$. What is the distance from $O$ to $A$?

[asy] usepackage("mathptmx"); size(6cm); defaultpen(linewidth(0.7)); real r = 1 - pi + sqrt(pi^2 + 1); pair O = (0, 0), A = r * dir(-30), B = (r + 2pi) * dir(-30), C = (r + 2pi) * dir(30), D = r * dir(30); draw(arc(O, r, 30, 330)); draw(arc(O, r + 2pi, -30, 30)); draw(O--A, dotted); draw(O--D, dotted); draw(A--B); draw(C--D); dot("$O$", O, dir(135)); dot("$A$", A, dir(270)); dot("$B$", B, dir(0)); dot("$C$", C, dir(0)); dot("$D$", D, 1.5 * dir(70)); [/asy]

$\textbf{(A)}~1 \qquad \textbf{(B)}~1 - \pi + \sqrt{\pi^{2} + 1} \qquad \textbf{(C)}~\frac{\pi}{2} \qquad \textbf{(D)}~\frac{\sqrt{\pi^{2} + 1}}{2} \qquad \textbf{(E)}~2$

Solution 1

Let the length of $OA=r$, which is the radius of the smaller circle. Then, the radius of the larger circle, $OB$, is equal to $r+2\pi$. Indeed, we know that the length of major arc $\widehat{AD}=2\pi$ and the length of minor arc $\widehat{BC}=2\pi$. So, using the formula for length of an arc formed by the central angle $\angle COB$, which we denote as $\theta$, we have that:

\[\widehat{AD}=2\pi=(2\pi-\theta) \cdot r,\] \[\widehat{BC}=2\pi=\theta \cdot (2\pi+r).\]

Expanding, we have \[\theta \cdot r +2 \pi \theta = 2\pi,\] \[2 \pi r-\theta r=2\pi,\] and by adding the two equations we have that

\[2 \pi \theta + 2 \pi r =4\pi \implies \theta + r=2.\]

Indeed, the question is asking for us to solve for $r$, and so we use $\theta = 2-r$ back into our original equation to solve:

\[(2 \pi - \theta) \cdot r = (2 \pi - 2 +r) \cdot r = (2 \pi -2)r +r^2= 2\pi \implies\] \[r^2+(2 \pi -2)r - 2 \pi =0.\]

Using the quadratic formula, we have that \[r=\frac{2-2\pi \pm \sqrt{(2 \pi -2)^2 - 4(1)(-2\pi)}}{2}=\frac{2-2\pi \pm \sqrt{4 \pi^2 -8 \pi +4+8 \pi}}{2}= \frac{2-2\pi \pm 2 \sqrt{1+ \pi^2}}{2}= 1- \pi \pm \sqrt{1+\pi^2}.\]

Since length must be positive, we consider only the positive root, and so the answer is $\boxed{\textbf{(B)} 1 - \pi + \sqrt{1 +\pi^2}}$.

~e_is_2.71828

See Also

2025 AMC 12A (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

These problems are copyrighted © by the Mathematical Association of America.

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