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

Revision as of 18:57, 15 October 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 $m+n$ .

$[asy] pair A = (2,1); pair B = (0,0); pair C = (3,0); dot(A^^B^^C); label("$A$", A, N); label("$B$", B, S); label("$C$", C, S); draw(A--B--C--cycle); for(real i=0.62; i<2.7; i+=0.29){ draw(circle((i,0.145), 0.145)); } [/asy]$

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 $x = \cot{\frac{B}{2}} + \cot{\frac{C}{2}}$ . 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

Solution 4

First, let the circle tangent to $AB$ and $BC$ be $O$ and the other circle that is tangent to $AC$ and $BC$ be $R$ . Let $x$ be the distance from the tangency point on line segment $BC$ of the circle $O$ to $B$ . Also, let $y$ be the distance of the tangency point of circle $R$ on the line segment $BC$ to point $C$ . Realize that we can let $n$ be the number of circles tangent to line segment $BC$ and $r$ be the corresponding radius of each of the circles. Also, the circles that are tangent to $BC$ are similar. So, we can build the equation $BC = (x+y+2(n-1)) \times r$ . Looking at the given information, we see that when $n=8$ , $r=34$ , and when $n=2024$ , $r=1$ , and we also want to find the radius $r$ in the case where $n=1$ . Using these facts, we can write the following equations:

$BC = (x+y+2(8-1)) \times 34 = (x+y+2(2024-1)) \times 1 = (x+y+2(1-1)) \times r$

We can find that $x+y = \frac{1190}{11}$ . Now, let $(x+y+2(2024-1)) \times 1 = (x+y+2(1-1)) \times r$ .

Substituting $x+y = \frac{1190}{11}$ in, we find that $r = \frac{192}{5} \implies \boxed{197}.$

~EaZ_Shadow

Solution 5 (one variable)

Define $I, x_1, x_8, y_1, y_{2024}$ to be the incenter and centers of the first and last circles of the $8$ and $2024$ tangent circles to $BC,$ and define $r$ to be the inradius of triangle $\bigtriangleup ABC.$ We calculate $\overline{x_1x_8} = 34 \cdot 14$ and $\overline{y_1y_{2024}} = 1 \cdot 4046$ because connecting the center of the circles voids two extra radii.

We can easily see that $B, x_1, x_8,$ and $I$ are collinear, and the same follows for $C, y_1, y_2024,$ and $I$ (think angle bisectors).

We observe that triangles $\bigtriangleup I x_1 x_8$ and $\bigtriangleup I y_1 y_{2024}$ are similar, and therefore the ratio of the altitude to the base is the same, so we note

$\frac{\text{altitude}}{\text{base}} = \frac{r-34}{34\cdot 14} = \frac{r-1}{1\cdot 4046}.$

Solving yields $r = \frac{192}{5},$ so the answer is $192+5 = \boxed{197}.$

-spectraldragon8

Video Solution 1 by OmegaLearn.org

https://youtu.be/MWTf6Jr8UwU

Video Solution

https://youtu.be/qlrVguS79GA

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://www.youtube.com/watch?v=MWhLgPr-ZR8&t=716s&ab_channel=TheBeautyofMath (beautyofmath)

@@ Line 58: / Line 58: @@
 Substituting <math>x+y = \frac{1190}{11}</math> in, we find that <cmath>r = \frac{192}{5} \implies \boxed{197}.</cmath>
-~EaZ_Shadow376
+~EaZ_Shadow
 ==Solution 5 (one variable)==

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

Page

Toolbox

Search

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

Revision as of 18:57, 15 October 2024

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5 (one variable)

Video Solution 1 by OmegaLearn.org

Video Solution

Video Solution

See also