Difference between revisions of "2013 AMC 10B Problems/Problem 16"
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Let us use mass points: | Let us use mass points: | ||
Assign <math>B</math> mass <math>1</math>. Thus, because <math>E</math> is the midpoint of <math>AB</math>, <math>A</math> also has a mass of <math>1</math>. Similarly, <math>C</math> has a mass of <math>1</math>. <math>D</math> and <math>E</math> each have a mass of <math>2</math> because they are between <math>B</math> and <math>C</math> and <math>A</math> and <math>B</math> respectively. Note that the mass of <math>D</math> is twice the mass of <math>A</math>, so AP must be twice as long as <math>PD</math>. PD has length <math>2</math>, so <math>AP</math> has length <math>4</math> and <math>AD</math> has length <math>6</math>. Similarly, <math>CP</math> is twice <math>PE</math> and <math>PE=1.5</math>, so <math>CP=3</math> and <math>CE=4.5</math>. Now note that triangle <math>PED</math> is a <math>3-4-5</math> right triangle with the right angle <math>DPE</math>. This means that the quadrilateral <math>AEDC</math> is a kite. The area of a kite is half the product of the diagonals, <math>AD</math> and <math>CE</math>. Recall that they are <math>6</math> and <math>4.5</math> respectively, so the area of <math>AEDC</math> is <math>6*4.5/2=\boxed{\textbf{(B)} 13.5}</math> | Assign <math>B</math> mass <math>1</math>. Thus, because <math>E</math> is the midpoint of <math>AB</math>, <math>A</math> also has a mass of <math>1</math>. Similarly, <math>C</math> has a mass of <math>1</math>. <math>D</math> and <math>E</math> each have a mass of <math>2</math> because they are between <math>B</math> and <math>C</math> and <math>A</math> and <math>B</math> respectively. Note that the mass of <math>D</math> is twice the mass of <math>A</math>, so AP must be twice as long as <math>PD</math>. PD has length <math>2</math>, so <math>AP</math> has length <math>4</math> and <math>AD</math> has length <math>6</math>. Similarly, <math>CP</math> is twice <math>PE</math> and <math>PE=1.5</math>, so <math>CP=3</math> and <math>CE=4.5</math>. Now note that triangle <math>PED</math> is a <math>3-4-5</math> right triangle with the right angle <math>DPE</math>. This means that the quadrilateral <math>AEDC</math> is a kite. The area of a kite is half the product of the diagonals, <math>AD</math> and <math>CE</math>. Recall that they are <math>6</math> and <math>4.5</math> respectively, so the area of <math>AEDC</math> is <math>6*4.5/2=\boxed{\textbf{(B)} 13.5}</math> |
Revision as of 16:04, 27 March 2013
Problem
In triangle , medians and intersect at , , , and . What is the area of ?
Solution
Solution 1
Let us use mass points: Assign mass . Thus, because is the midpoint of , also has a mass of . Similarly, has a mass of . and each have a mass of because they are between and and and respectively. Note that the mass of is twice the mass of , so AP must be twice as long as . PD has length , so has length and has length . Similarly, is twice and , so and . Now note that triangle is a right triangle with the right angle . This means that the quadrilateral is a kite. The area of a kite is half the product of the diagonals, and . Recall that they are and respectively, so the area of is
Solution 2
Note that triangle is a right triangle, and that the four angles that have point are all right angles. Using the fact that the centroid () divides each median in a ratio, and . Quadrilateral is now just four right triangles. The area is
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |