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Difference between revisions of "2003 AIME II Problems/Problem 6"

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Therefore, <math>\left[\Delta ABC \bigcup \Delta A'B'C'\right] = \frac{12}{9}[\Delta ABC]= \frac{4}{3}\cdot84=\boxed{112}</math>.
 
Therefore, <math>\left[\Delta ABC \bigcup \Delta A'B'C'\right] = \frac{12}{9}[\Delta ABC]= \frac{4}{3}\cdot84=\boxed{112}</math>.
  
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==Solution 2==
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==Solution 2(Doesn’t require good diagram)==
  
 
First, find the area of <math>\Delta ABC</math> either like the first solution or by using Heron’s Formula. Then, draw the medians from <math>G</math> to each of <math>A, B, C, A’, B’,</math> and <math>C’</math>. Since the medians of a triangle divide the triangle into 6 triangles with equal area, we can find that each of the 6 outer triangles have equal area. (Proof: Since I’m too lazy to draw out a diagram, I’ll just have you borrow the one above. Draw medians <math>GA</math> and <math>GB’</math>, and let’s call the points that <math>GA</math> intersects <math>C’B’</math> “<math>H</math>” and the point <math>GB’</math> intersects <math>AC</math> “<math>I</math>”. From the previous property and the fact that both <math>\Delta ABC</math> and <math>\Delta A’B’C’</math> are congruent, <math>\Delta GHB’</math> has the same area as <math>\Delta GIA</math>. Because of that, both “half” triangles created also have the same area. The same logic can be applied to all other triangles).
 
First, find the area of <math>\Delta ABC</math> either like the first solution or by using Heron’s Formula. Then, draw the medians from <math>G</math> to each of <math>A, B, C, A’, B’,</math> and <math>C’</math>. Since the medians of a triangle divide the triangle into 6 triangles with equal area, we can find that each of the 6 outer triangles have equal area. (Proof: Since I’m too lazy to draw out a diagram, I’ll just have you borrow the one above. Draw medians <math>GA</math> and <math>GB’</math>, and let’s call the points that <math>GA</math> intersects <math>C’B’</math> “<math>H</math>” and the point <math>GB’</math> intersects <math>AC</math> “<math>I</math>”. From the previous property and the fact that both <math>\Delta ABC</math> and <math>\Delta A’B’C’</math> are congruent, <math>\Delta GHB’</math> has the same area as <math>\Delta GIA</math>. Because of that, both “half” triangles created also have the same area. The same logic can be applied to all other triangles).
 
Also, since the centroid of a triangle divides each median with the ratio <math>2:1</math>, along with the previous fact, each outer triangle has <math>1/9</math> the area of <math>\Delta ABC</math> and <math>\Delta A’B’C’</math>. Thus, the area of the region required is <math>\frac{4}{3}</math> times the area of <math>\Delta ABC</math> which is <math>\boxed {112}</math>.
 
Also, since the centroid of a triangle divides each median with the ratio <math>2:1</math>, along with the previous fact, each outer triangle has <math>1/9</math> the area of <math>\Delta ABC</math> and <math>\Delta A’B’C’</math>. Thus, the area of the region required is <math>\frac{4}{3}</math> times the area of <math>\Delta ABC</math> which is <math>\boxed {112}</math>.
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Solution by Someonenumber011
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==Solution 3 (Rigorous)==
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<math>[ABC]</math> can be calculated as 84 using Heron's formula or other methods. Since a <math>180^{\circ}</math> rotation is equivalent to reflection through a point, we have a homothety with scale factor <math>-1</math> from <math>ABC</math> to <math>A'B'C'</math> through the centroid <math>G</math>. Let <math>M</math> be the midpoint of <math>BC</math> which maps to <math>M'</math> and note that <math>A'G=AG=2GM,</math> implying that <math>GM=MA'.</math> Similarly, we have <math>AM'=M'G=GM=MA'.</math> Also let <math>D</math> and <math>E</math> be the intersections of <math>BC</math> with <math>A'B'</math> and <math>A'C',</math> respectively. The homothety implies that we must have <math>DE || B'C',</math> so there is in fact another homothety centered at <math>A'</math> taking <math>A'DE</math> to <math>A'B'C'</math>. Since <math>A'M'=3A'M,</math> the scale factor of this homothety is 3 and thus <math>[A'DE]=\frac{1}{9}[A'B'C']=\frac{1}{9}[ABC].</math> We can apply similar reasoning to the other small triangles in <math>A'B'C'</math> not contained within <math>ABC</math>, so our final answer is <math>[ABC]+3\cdot\frac{1}{9}[ABC]=\boxed{112.}</math>
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==Video Solution by Sal Khan==
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https://www.youtube.com/watch?v=l9j26EOvTYc&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=17 - AMBRIGGS
  
 
== See also ==
 
== See also ==

Latest revision as of 16:38, 30 July 2022

Problem

In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$

Solution

Since a $13-14-15$ triangle is a $5-12-13$ triangle and a $9-12-15$ triangle "glued" together on the $12$ side, $[ABC]=\frac{1}{2}\cdot12\cdot14=84$.

There are six points of intersection between $\Delta ABC$ and $\Delta A'B'C'$. Connect each of these points to $G$.

[asy] size(8cm); pair A,B,C,G,D,E,F,A_1,A_2,B_1,B_2,C_1,C_2; B=(0,0); A=(5,12); C=(14,0); E=(12.6667,8); D=(7.6667,-4); F=(-1.3333,8); G=(6.3333,4); B_1=(4.6667,0); B_2=(1.6667,4); A_1=(3.3333,8); A_2=(8,8); C_1=(11,4); C_2=(9.3333,0); dot(A); dot(B); dot(C); dot(G); dot(D); dot(E); dot(F); dot(A_1); dot(B_1); dot(C_1); dot(A_2); dot(B_2); dot(C_2); draw(B--A--C--cycle); draw(E--D--F--cycle); draw(B_1--A_2); draw(A_1--C_2); draw(C_1--B_2); label("$B$",B,WSW); label("$A$",A,N); label("$C$",C,ESE); label("$G$",G,S); label("$B'$",E,ENE); label("$A'$",D,S); label("$C'$",F,WNW); [/asy]

There are $12$ smaller congruent triangles which make up the desired area. Also, $\Delta ABC$ is made up of $9$ of such triangles. Therefore, $\left[\Delta ABC \bigcup \Delta A'B'C'\right] = \frac{12}{9}[\Delta ABC]= \frac{4}{3}\cdot84=\boxed{112}$.

Solution 2(Doesn’t require good diagram)

First, find the area of $\Delta ABC$ either like the first solution or by using Heron’s Formula. Then, draw the medians from $G$ to each of $A, B, C, A’, B’,$ and $C’$. Since the medians of a triangle divide the triangle into 6 triangles with equal area, we can find that each of the 6 outer triangles have equal area. (Proof: Since I’m too lazy to draw out a diagram, I’ll just have you borrow the one above. Draw medians $GA$ and $GB’$, and let’s call the points that $GA$ intersects $C’B’$ “$H$” and the point $GB’$ intersects $AC$ “$I$”. From the previous property and the fact that both $\Delta ABC$ and $\Delta A’B’C’$ are congruent, $\Delta GHB’$ has the same area as $\Delta GIA$. Because of that, both “half” triangles created also have the same area. The same logic can be applied to all other triangles). Also, since the centroid of a triangle divides each median with the ratio $2:1$, along with the previous fact, each outer triangle has $1/9$ the area of $\Delta ABC$ and $\Delta A’B’C’$. Thus, the area of the region required is $\frac{4}{3}$ times the area of $\Delta ABC$ which is $\boxed {112}$.

Solution by Someonenumber011

Solution 3 (Rigorous)

$[ABC]$ can be calculated as 84 using Heron's formula or other methods. Since a $180^{\circ}$ rotation is equivalent to reflection through a point, we have a homothety with scale factor $-1$ from $ABC$ to $A'B'C'$ through the centroid $G$. Let $M$ be the midpoint of $BC$ which maps to $M'$ and note that $A'G=AG=2GM,$ implying that $GM=MA'.$ Similarly, we have $AM'=M'G=GM=MA'.$ Also let $D$ and $E$ be the intersections of $BC$ with $A'B'$ and $A'C',$ respectively. The homothety implies that we must have $DE || B'C',$ so there is in fact another homothety centered at $A'$ taking $A'DE$ to $A'B'C'$. Since $A'M'=3A'M,$ the scale factor of this homothety is 3 and thus $[A'DE]=\frac{1}{9}[A'B'C']=\frac{1}{9}[ABC].$ We can apply similar reasoning to the other small triangles in $A'B'C'$ not contained within $ABC$, so our final answer is $[ABC]+3\cdot\frac{1}{9}[ABC]=\boxed{112.}$

Video Solution by Sal Khan

https://www.youtube.com/watch?v=l9j26EOvTYc&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=17 - AMBRIGGS

See also

2003 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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