Difference between revisions of "2024 AIME I Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | 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 BC <math>a and b</math>. Now we have the length of side BC of being <math>(2)(2022)+1+1+a+b</math>. However, the side BC can also be written as <math>(6)(68)+34+34+34a+34b</math>, due to similar triangles from the second diagram. If we set the equations equal, we have <math>\frac{1190}{11} = a+b</math>. Call the radius of the incircle r, then we have the side BC to be <math>r(a+b)</math>. We find <math>r</math> as <math>\frac{4046+\frac{1190}{11}}{\frac{1190}{11}}</math>, which simplifies to <math>\frac{10+((34)(11))}{10}</math>,so we have <math>\frac{192}{5}</math>, which sums to 197. | + | 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 BC <math>a</math> and <math>b</math>. Now we have the length of side BC of being <math>(2)(2022)+1+1+a+b</math>. However, the side BC can also be written as <math>(6)(68)+34+34+34a+34b</math>, due to similar triangles from the second diagram. If we set the equations equal, we have <math>\frac{1190}{11} = a+b</math>. Call the radius of the incircle r, then we have the side BC to be <math>r(a+b)</math>. We find <math>r</math> as <math>\frac{4046+\frac{1190}{11}}{\frac{1190}{11}}</math>, which simplifies to <math>\frac{10+((34)(11))}{10}</math>,so we have <math>\frac{192}{5}</math>, which sums to 197. |
==See also== | ==See also== | ||
Revision as of 16:39, 2 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 
.
Solution
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 BC 
and 
. Now we have the length of side BC of being 
. However, the side BC 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 r, then we have the side BC to be 
. We find 
as 
, which simplifies to 
,so we have 
, which sums to 197.
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. 
