Difference between revisions of "2003 AIME II Problems/Problem 6"
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== 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>. | ||
== See also == | == See also == | ||
{{AIME box|year=2003|n=II|num-b=5|num-a=7}} | {{AIME box|year=2003|n=II|num-b=5|num-a=7}} | ||
Revision as of 18:28, 26 July 2008
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 if 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, ![$[ABC]=\frac{1}{2}\cdot12\cdot14=84$](http://latex.artofproblemsolving.com/c/2/3/c23eb021f542e6ca0c51b59e044a8d72a3644a86.png)
.
There are six points of intersection between 
and 
. Connect each of these points to .
![[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]](http://latex.artofproblemsolving.com/a/5/d/a5d42315a7709b2d03b4403cc072e4b6c1a6fd51.png)
There are smaller congruent triangles which make up the desired area. Also, is made up of 
of such triangles.
Therefore, ![$\left[\Delta ABC \bigcup \Delta A'B'C'\right] = \frac{12}{9}[\Delta ABC]= \frac{4}{3}\cdot84=\boxed{112}$](http://latex.artofproblemsolving.com/c/e/6/ce6302862d09f736b0fabb7cb16d47f75107fba7.png)
.
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 | ||
