Difference between revisions of "2024 AIME I Problems/Problem 8"
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==Solution 3== | ==Solution 3== | ||
− | Let <math>x = \cot{\frac{ | + | Let <math>x = \cot{\frac{B}{2}} + \cot{\frac{C}{2}}</math>. By representing <math>BC</math> in two ways, we have the following: |
<cmath>34x + 7\cdot 34\cdot 2 = BC</cmath> | <cmath>34x + 7\cdot 34\cdot 2 = BC</cmath> | ||
<cmath>x + 2023 \cdot 2 = BC</cmath> | <cmath>x + 2023 \cdot 2 = BC</cmath> | ||
Line 47: | Line 47: | ||
Thus <cmath>r = \frac{192}{5} \implies \boxed{197}.</cmath> | Thus <cmath>r = \frac{192}{5} \implies \boxed{197}.</cmath> | ||
~AtharvNaphade | ~AtharvNaphade | ||
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==Solution 4== | ==Solution 4== |
Revision as of 22:04, 3 February 2024
Contents
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 .
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 .
~eevee9406
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
Solution 4
First, let the circle tangent to and be and the other circle that is tangent to and be . Let be the distance from the tangency point on line segment of the circle to . Also, let be the distance of the tangency point of circle on the line segment to point . Realize that we can let be the number of circles tangent to line segment and be the corresponding radius of each of the circles. Also, the circles that are tangent to are similar. So, we can build the equation . Looking at the given information, we see that when , , and when , , and we also want to find the radius in the case where . Using these facts, we can write the following equations:
We can find that . Now, let .
Substituting in, we find that
~Rainier2020
Solution 5 (one variable)
Define to be the incenter and centers of the first and last circles of the and tangent circles to and define to be the inradius of triangle We calculate and because connecting the center of the circles voids two extra radii.
We can easily see that and are collinear, and the same follows for and (think angle bisectors).
We observe that triangles and are similar, and therefore the ratio of the altitude to the base is the same, so we note
Solving yields so the answer is
-spectraldragon8
Video Solution 1 by OmegaLearn.org
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.