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>. | ||
− | ==Solution 2== | + | ==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). |
Revision as of 15:12, 28 December 2019
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
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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