Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Video Solution（Chinese）subtitle in English)
 
(17 intermediate revisions by 15 users not shown)
Line 2: Line 2:
 
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 <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>.
  
−
==Solution==
+
<asy>
−
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.
+
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 <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>.
 +
 
 +
==Solution 2==
 +
Assume that <math>ABC</math> is isosceles with <math>AB=AC</math>.
 +
 
 +
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.
 +
 
 +
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>.
 +
 
 +
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:
 +
 
 +
<math>r=1+\frac{2023}{x}</math>
 +
 
 +
<math>=1+\frac{11\cdot2023}{595}</math>
 +
 
 +
<math>=1+\frac{187}{5}</math>
 +
 
 +
<math>=\frac{192}{5}</math>
 +
 
 +
Thus the answer is <math>192+5=\boxed{197}</math>.
 +
 
 +
~eevee9406
 +
 
 +
==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:
 +
<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
 +
 
 +
==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:
 +
 
 +
<math>BC = (x+y+2(8-1)) \times 34 = (x+y+2(2024-1)) \times 1 = (x+y+2(1-1)) \times r</math>
 +
 
 +
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>.
 +
 
 +
Substituting <math>x+y = \frac{1190}{11}</math> in, we find that <cmath>r = \frac{192}{5} \implies \boxed{197}.</cmath>
 +
 
 +
~EaZ_Shadow
 +
 
 +
==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.
 +
 
 +
 
 +
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 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
 +
 
 +
<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>
 +
 
 +
-[https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 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==
 
==See also==

Latest revision as of 03:32, 25 December 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（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
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AIME_I_Problems/Problem_8&oldid=237790"