2025 AMC 12A Problems/Problem 10: Difference between revisions
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==See Also== | |||
{{AMC10 box|year=2025|ab=A|before=First Problem|num-a=2}} | |||
{{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}} | |||
Revision as of 16:13, 6 November 2025
Problem
In the figure shown below, major arc 
and minor arc 
have the same center, 
. Also, 
lies between and 
, and 
lies between and 
. Major arc , minor arc , and each of the two segments 
and 
have length 
. What is the distance from to ?
![[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]](http://latex-new.aopstest.com/9/0/f/90fbb050b92f4409fa22f130e65b4a4036650ce9.png)
Solution 1
Let the length of 
, which is the radius of the smaller circle. Then, the radius of the larger circle, 
, is equal to 
. Indeed, we know that the length of major arc 
and the length of minor arc 
. So, using the formula for length of an arc formed by the central angle 
, which we denote as 
, we have that:
![\[\widehat{AD}=2\pi=(2\pi-\theta) \cdot r,\]](http://latex-new.aopstest.com/3/b/f/3bfa6026e2848fb3951574bef7b50fa324c6cea4.png)
![\[\widehat{BC}=2\pi=\theta \cdot (2\pi+r).\]](http://latex-new.aopstest.com/7/a/5/7a58860c9530564bcae5605dbd31717ff98ea776.png)
Expanding, we have
![\[\theta \cdot r +2 \pi \theta = 2\pi,\]](http://latex-new.aopstest.com/6/5/e/65e0af9274c41bfbb4ac72422fc3ba63bb322d9e.png)
![\[2 \pi r-\theta r=2\pi,\]](http://latex-new.aopstest.com/9/8/4/9842f8214a090e5b5988979365f302dbfca13ba8.png)
and by adding the two equations we have that
![\[2 \pi \theta + 2 \pi r =4\pi \implies \theta + r=2.\]](http://latex-new.aopstest.com/8/c/c/8cc1a67856f05e6860a95cf603d32bf339b372ae.png)
Indeed, the question is asking for us to solve for 
, and so we use 
back into our original equation to solve:
![\[(2 \pi - \theta) \cdot r = (2 \pi - 2 +r) \cdot r = (2 \pi -2)r +r^2= 2\pi \implies\]](http://latex-new.aopstest.com/d/4/1/d4135acf2667056ab13a6d301d501ad44850d9a3.png)
![\[r^2+(2 \pi -2)r - 2 \pi =0.\]](http://latex-new.aopstest.com/2/e/8/2e8e6c67f29aa2fc461eb00678c2a929267c45e4.png)
Using the quadratic formula, we have that
![\[r=\frac{2-2\pi \pm \sqrt{(2 \pi -2)^2 - 4(1)(-2\pi)}}{2}=\frac{2-2\pi \pm \sqrt{4 \pi^2 -8 \pi +4+8 \pi}}{2}= \frac{2-2\pi \pm 2 \sqrt{1+ \pi^2}}{2}= 1- \pi \pm \sqrt{1+\pi^2}.\]](http://latex-new.aopstest.com/0/4/4/044fa4acfcd3304a010c260f0f5cefbb006259a9.png)
Since length must be positive, we consider only the positive root, and so the answer is 
.
~e_is_2.71828
See Also
| 2025 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 10 Problems and Solutions | ||
| 2025 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 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. 
