Difference between revisions of "2003 AIME II Problems/Problem 6"
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== Problem == | == Problem == | ||
+ | In triangle <math>ABC,</math> <math>AB = 13,</math> <math>BC = 14,</math> <math>AC = 15,</math> and point <math>G</math> is the intersection of the medians. Points <math>A',</math> <math>B',</math> and <math>C',</math> are the images of <math>A,</math> <math>B,</math> and <math>C,</math> respectively, after a <math>180^\circ</math> rotation about <math>G.</math> What is the area of the union of the two regions enclosed by the triangles <math>ABC</math> and <math>A'B'C'?</math> | ||
== Solution == | == Solution == | ||
+ | |||
+ | Since a <math>13-14-15</math> triangle is a <math>5-12-13</math> triangle and a <math>9-12-15</math> triangle "glued" together on the <math>12</math> side, <math>[ABC]=\frac{1}{2}\cdot12\cdot14=84</math>. | ||
+ | |||
+ | There are six points of intersection between <math>\Delta ABC</math> and <math>\Delta A'B'C'</math>. Connect each of these points to <math>G</math>. | ||
+ | |||
+ | <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 <math>12</math> smaller congruent triangles which make up the desired area. Also, <math>\Delta ABC</math> is made up of <math>9</math> of such triangles. | ||
+ | Therefore, <math>\left[\Delta ABC \bigcup \Delta A'B'C'\right] = \frac{12}{9}[\Delta ABC]= \frac{4}{3}\cdot84=\boxed{112}</math>. | ||
+ | |||
+ | ==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). | ||
+ | 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>. | ||
+ | |||
+ | Solution by Someonenumber011 | ||
+ | |||
+ | ==Solution 3 (Rigorous)== | ||
+ | |||
+ | <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> | ||
+ | |||
+ | ==Video Solution by Sal Khan== | ||
+ | https://www.youtube.com/watch?v=l9j26EOvTYc&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=17 - AMBRIGGS | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2003|n=II|num-b=5|num-a=7}} | |
+ | |||
+ | [[Category: Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:38, 30 July 2022
Contents
Problem
In triangle
and point
is the intersection of the medians. Points
and
are the images of
and
respectively, after a
rotation about
What is the area of the union of the two regions enclosed by the triangles
and
Solution
Since a triangle is a
triangle and a
triangle "glued" together on the
side,
.
There are six points of intersection between and
. Connect each of these points to
.
There are smaller congruent triangles which make up the desired area. Also,
is made up of
of such triangles.
Therefore,
.
Solution 2(Doesn’t require good diagram)
First, find the area of either like the first solution or by using Heron’s Formula. Then, draw the medians from
to each of
and
. 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
and
, and let’s call the points that
intersects
“
” and the point
intersects
“
”. From the previous property and the fact that both
and
are congruent,
has the same area as
. 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
, along with the previous fact, each outer triangle has
the area of
and
. Thus, the area of the region required is
times the area of
which is
.
Solution by Someonenumber011
Solution 3 (Rigorous)
can be calculated as 84 using Heron's formula or other methods. Since a
rotation is equivalent to reflection through a point, we have a homothety with scale factor
from
to
through the centroid
. Let
be the midpoint of
which maps to
and note that
implying that
Similarly, we have
Also let
and
be the intersections of
with
and
respectively. The homothety implies that we must have
so there is in fact another homothety centered at
taking
to
. Since
the scale factor of this homothety is 3 and thus
We can apply similar reasoning to the other small triangles in
not contained within
, so our final answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.