Difference between revisions of "2024 AIME I Problems/Problem 8"
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+ | ==Solution 3== | ||
+ | |||
+ | Let <math>x = \cot{\frac{A}{2}} + \cot{\frac{B}{2}}</math>. By representing <math>BC</math> in two ways, we have the following: | ||
+ | <cmath>34x + 7\cdot 34\cdot 2 = BC</cmath> | ||
+ | <cmath>x + 2023 \cdot 2 = BC</cmath> | ||
+ | |||
+ | Solving we find <math>x = \frac{1190}{11}</math>. | ||
+ | Now draw the inradius, let it be <math>r</math>. We find that <math>rx =BC</math>, hence | ||
+ | <cmath>xr = x + 4046 \implies r-1 = \frac{11}{1190}\cdot 4046 = \frac{187}{5}.</cmath> | ||
+ | Thus <cmath>r = \frac{192}{5} \implies \boxed{197}.</cmath> | ||
+ | ~AtharvNaphade | ||
==See also== | ==See also== |
Revision as of 14:38, 3 February 2024
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 .
~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
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.