Difference between revisions of "2023 AMC 12B Problems/Problem 3"
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+ | {{duplicate|[[2023 AMC 10B Problems/Problem 3|2023 AMC 10B #3]] and [[2023 AMC 12B Problems/Problem 3|2023 AMC 12B #3]]}} | ||
==Problem== | ==Problem== | ||
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==Solution 1== | ==Solution 1== | ||
− | Because the triangle are right triangles, we know the hypotenuses are diameters of circles <math>A</math> and <math>B</math>. Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply <math>\pi</math> to get <math>6.25\pi</math> and <math>42.25\pi</math> as the areas of the circles. Multiply 4 on both numbers to get <math>25\pi</math> and <math>169\pi</math>. Cancel out the <math>\pi</math>, and lastly, divide, to get your answer <math>\boxed{\textbf(D)\frac{25}{169}}</math>. | + | Because the triangle are right triangles, we know the hypotenuses are diameters of circles <math>A</math> and <math>B</math>. Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply <math>\pi</math> to get <math>6.25\pi</math> and <math>42.25\pi</math> as the areas of the circles. Multiply 4 on both numbers to get <math>25\pi</math> and <math>169\pi</math>. Cancel out the <math>\pi</math>, and lastly, divide, to get your answer <math>\boxed{\textbf(D)\frac{25}{169}}</math>. |
~Failure.net | ~Failure.net | ||
+ | |||
+ | ==Solution 2== | ||
+ | Since the arc angle of the diameter of a circle is <math>90</math> degrees, the hypotenuse of each these two triangles is respectively the diameter of circles <math>A</math> and <math>B</math>. | ||
+ | |||
+ | Therefore the ratio of the areas equals the radius of circle <math>A</math> squared : the radius of circle <math>B</math> squared | ||
+ | <math>=</math> <math>0.5\times</math> the diameter of circle <math>A</math>, squared : <math>0.5\times</math> the diameter of circle <math>B</math>, squared | ||
+ | <math>=</math> the diameter of circle <math>A</math>, squared: the diameter of circle <math>B</math>, squared <math>=\boxed{\textbf{(D) }\frac{25}{169}}.</math> | ||
+ | |||
+ | |||
+ | ~Mintylemon66 | ||
+ | |||
+ | ==Solution 3== | ||
+ | The ratio of areas of circles is the same as the ratios of the diameters squared. Since this is a right triangle the hypotenuse of each triangle will be the diameter of the circle. This yields the expression <math>\frac{5^2}{13^2} =\boxed{\textbf{(D) }\frac{25}{169}}.</math> | ||
+ | |||
+ | ~vsinghminhas | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC10 box|year=2023|ab=B|num-b=2|num-a=4}} | ||
+ | {{AMC12 box|year=2023|ab=B|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Revision as of 19:19, 15 November 2023
- The following problem is from both the 2023 AMC 10B #3 and 2023 AMC 12B #3, so both problems redirect to this page.
Contents
[hide]Problem
A 3-4-5 right triangle is inscribed circle , and a 5-12-13 right triangle is inscribed in circle . What is the ratio of the area of circle to circle ?
Solution 1
Because the triangle are right triangles, we know the hypotenuses are diameters of circles and . Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply to get and as the areas of the circles. Multiply 4 on both numbers to get and . Cancel out the , and lastly, divide, to get your answer .
~Failure.net
Solution 2
Since the arc angle of the diameter of a circle is degrees, the hypotenuse of each these two triangles is respectively the diameter of circles and .
Therefore the ratio of the areas equals the radius of circle squared : the radius of circle squared the diameter of circle , squared : the diameter of circle , squared the diameter of circle , squared: the diameter of circle , squared
~Mintylemon66
Solution 3
The ratio of areas of circles is the same as the ratios of the diameters squared. Since this is a right triangle the hypotenuse of each triangle will be the diameter of the circle. This yields the expression
~vsinghminhas
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 |
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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.