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2024 AIME I Problems/Problem 8: Difference between revisions

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


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


==Solution 2==
==Solution 2==
Assume that <math>ABC</math> is isosceles with <math>AB=AC</math>.
Assume that <imath>ABC</imath> is isosceles with <imath>AB=AC</imath>.


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


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


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


<math>r=1+\frac{2023}{x}</math>
<cmath>r=1+\frac{2023}{x}=1+\frac{11\cdot2023}{595}=1+\frac{187}{5}=\frac{192}{5}</cmath>


<math>=1+\frac{11\cdot2023}{595}</math>
Thus the answer is <imath>192+5=\boxed{197}</imath>.
 
<math>=1+\frac{187}{5}</math>
 
<math>=\frac{192}{5}</math>
 
Thus the answer is <math>192+5=\boxed{197}</math>.


~eevee9406
~eevee9406
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==Solution 3==
==Solution 3==


Let <math>x = \cot{\frac{B}{2}} + \cot{\frac{C}{2}}</math>. By representing <math>BC</math> in two ways, we have the following:
Let <imath>x = \cot{\frac{B}{2}} + \cot{\frac{C}{2}}</imath>. By representing <imath>BC</imath> in two ways, we have the following:
<cmath>34x + 7\cdot 34\cdot 2 = BC</cmath>
<cmath>34x + 7\cdot 34\cdot 2 = BC</cmath>
<cmath>x + 2023 \cdot 2 = BC</cmath>
<cmath>x + 2023 \cdot 2 = BC</cmath>


Solving we find <math>x = \frac{1190}{11}</math>.  
Solving we find <imath>x = \frac{1190}{11}</imath>.  
Now draw the inradius, let it be <math>r</math>. We find that <math>rx =BC</math>, hence  
Now draw the inradius, let it be <imath>r</imath>. We find that <imath>rx =BC</imath>, hence  
<cmath>xr = x + 4046 \implies r-1 = \frac{11}{1190}\cdot 4046 = \frac{187}{5}.</cmath>
<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>
Thus <cmath>r = \frac{192}{5} \implies \boxed{197}.</cmath>
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==Solution 4==
==Solution 4==


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


<math>BC = (x+y+2(8-1)) \times 34 = (x+y+2(2024-1)) \times 1 = (x+y+2(1-1)) \times r</math>
<imath>BC = (x+y+2(8-1)) \times 34 = (x+y+2(2024-1)) \times 1 = (x+y+2(1-1)) \times r</imath>


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


Substituting <math>x+y = \frac{1190}{11}</math> in, we find that <cmath>r = \frac{192}{5} \implies \boxed{197}.</cmath>
Substituting <imath>x+y = \frac{1190}{11}</imath> in, we find that <cmath>r = \frac{192}{5} \implies \boxed{197}.</cmath>


~EaZ_Shadow
~EaZ_Shadow
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==Solution 5 (one variable)==
==Solution 5 (one variable)==


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




We can easily see that <math>B, x_1, x_8,</math> and <math>I</math> are collinear, and the same follows for <math>C, y_1, y_2024,</math> and <math>I</math> (think angle bisectors).
We can easily see that <imath>B, x_1, y_1,</imath> and <imath>I</imath> are collinear, and the same follows for <imath>C, x_8, y_{2024},</imath> and <imath>I</imath> (think angle bisectors).




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


<cmath>\frac{\text{altitude}}{\text{base}} = \frac{r-34}{34\cdot 14} = \frac{r-1}{1\cdot 4046}.</cmath>
<cmath>\frac{\text{altitude}}{\text{base}} = \frac{r-34}{34\cdot 14} = \frac{r-1}{1\cdot 4046}.</cmath>




Solving yields <math>r = \frac{192}{5},</math> so the answer is <math>192+5 = \boxed{197}.</math>
Solving yields <imath>r = \frac{192}{5},</imath> so the answer is <imath>192+5 = \boxed{197}.</imath>


-[https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8]
-[https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8]


==Video Solution（国语）subtitle in English==
==Video Solution（Chinese）subtitle in English==
https://youtu.be/q8N-zzlUFpA
https://youtu.be/q8N-zzlUFpA


Latest revision as of 23:36, 11 November 2025

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 $B, x_1, y_1,$ and $I$ are collinear, and the same follows for $C, x_8, y_{2024},$ 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（Chinese）subtitle in English

https://youtu.be/q8N-zzlUFpA

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)

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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