2024 AIME I Problems/Problem 8
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[hide]Problem
Eight circles of radius are sequentially tangent, and two of the circles are tangent to and of triangle , respectively. circles of radius can be arranged in the same manner. The inradius of triangle can be expressed as , where and are relatively prime positive integers. Find .
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Solution 1
Draw an altitude from both end circles of the diagram with the circles of radius one, and call the lengths you get drawing the altitudes of the circles down to and . Now we have the length of side of being . However, the side can also be written as , due to similar triangles from the second diagram. If we set the equations equal, we have . Call the radius of the incircle , then we have the side BC to be . We find as , which simplifies to ,so we have , which sums to .
Solution 2
Assume that is isosceles with .
If we let be the intersection of and the leftmost of the eight circles of radius , the center of the leftmost circle, and the intersection of the leftmost circle and , and we do the same for the circles of radius , naming the points , , and , respectively, then we see that . The same goes for vertex , and the corresponding quadrilaterals are congruent.
Let . We see that by similarity ratios (due to the radii). The corresponding figures on vertex are also these values. If we combine the distances of the figures, we see that and , and solving this system, we find that .
If we consider that the incircle of is essentially the case of circle with radius (the inradius of , we can find that . From , we have:
Thus the answer is .
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Solution 3
Let . By representing in two ways, we have the following:
Solving we find . Now draw the inradius, let it be . We find that , hence Thus ~AtharvNaphade
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See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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