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

Revision as of 13:38, 3 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 $m+n$ .

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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 $BC$ $a$ 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

Solution 3

Let $x = \cot{\frac{A}{2}} + \cot{\frac{B}{2}}$ . By representing $BC$ in two ways, we have the following: $34x + 7\cdot 34\cdot 2 = BC$ $x + 2023 \cdot 2 = BC$

Solving we find $x = \frac{1190}{11}$ . Now draw the inradius, let it be $r$ . We find that $rx =BC$ , hence $xr = x + 4046 \implies r-1 = \frac{11}{1190}\cdot 4046 = \frac{187}{5}.$ Thus $r = \frac{192}{5} \implies \boxed{197}.$ ~AtharvNaphade

@@ Line 27: / Line 27: @@
 ~eevee9406
+==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==

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
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