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Difference between revisions of "2024 AIME I Problems/Problem 8"

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Thus the answer is <math>192+5=\boxed{197}</math>.
 
Thus the answer is <math>192+5=\boxed{197}</math>.
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~eevee9406
  
 
==See also==
 
==See also==

Revision as of 19:20, 2 February 2024

Problem

Eight circles of radius $34$ are sequentially tangent, and two of the circles are tangent to $AB$ and $BC$ of triangle $ABC$, respectively. $2024$ circles of radius $1$ can be arranged in the same manner. The inradius of triangle $ABC$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

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 $BC$ $a$ and $b$. Now we have the length of side $BC$ of being $(2)(2022)+1+1+a+b$. However, the side $BC$ can also be written as $(6)(68)+34+34+34a+34b$, due to similar triangles from the second diagram. If we set the equations equal, we have $\frac{1190}{11} = a+b$. Call the radius of the incircle $r$, then we have the side BC to be $r(a+b)$. We find $r$ as $\frac{4046+\frac{1190}{11}}{\frac{1190}{11}}$, which simplifies to $\frac{10+((34)(11))}{10}$,so we have $\frac{192}{5}$, which sums to $\boxed{197}$.

Solution 2

Assume that $ABC$ is isosceles with $AB=AC$.

If we let $P_1$ be the intersection of $BC$ and the leftmost of the eight circles of radius $34$, $N_1$ the center of the leftmost circle, and $M_1$ the intersection of the leftmost circle and $AB$, and we do the same for the $2024$ circles of radius $1$, naming the points $P_2$, $N_2$, and $M_2$, respectively, then we see that $BP_1N_1M_1\sim BP_2N_2M_2$. The same goes for vertex $C$, and the corresponding quadrilaterals are congruent.

Let $x=BP_2$. We see that $BP_1=34x$ by similarity ratios (due to the radii). The corresponding figures on vertex $C$ are also these values. If we combine the distances of the figures, we see that $BC=2x+4046$ and $BC=68x+476$, and solving this system, we find that $x=\frac{595}{11}$.

If we consider that the incircle of $\triangle ABC$ is essentially the case of $1$ circle with $r$ radius (the inradius of $\triangle ABC$, we can find that $BC=2rx$. From $BC=2x+4046$, we have:

$r=1+\frac{2023}{x}$

$=1+\frac{11\cdot2023}{595}$

$=1+\frac{187}{5}$

$=\frac{192}{5}$

Thus the answer is $192+5=\boxed{197}$.

~eevee9406

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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