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Triangle and Midpoint Meet

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by avenus » Tue May 12, 2009 1:11 am
drawing the midpoints and using the properties of similar triangles, we can derive the vertices:

(-1,6), (3,-2), (7,2)

Answer is D

Note that by joining the midpoints of the sides of a triangle, you'll be effectively creating four congruent triangles.
Last edited by avenus on Tue May 12, 2009 1:50 am, edited 1 time in total.
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by Pranay » Tue May 12, 2009 1:28 am
Another way to solve the problem,

let the vertices be (x1,y1), (x2,y2), (x3,y3):

let the point (5,0) be the midpoint of (x1,y1) and (x2,y2)
|||y, (3,4) (x2, y2) and (x3,y3)
(1,2) (x1,y1) and (x3,y3)

Then applying the midpoint formula to all the midpoints and adding only the x-cordinates of will give the sum as 9.

Thus, D.

Hope it helps.
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by dtweah » Tue May 12, 2009 1:52 am
Pranay wrote:Another way to solve the problem,

let the vertices be (x1,y1), (x2,y2), (x3,y3):

let the point (5,0) be the midpoint of (x1,y1) and (x2,y2)
|||y, (3,4) (x2, y2) and (x3,y3)
(1,2) (x1,y1) and (x3,y3)

Then applying the midpoint formula to all the midpoints and adding only the x-cordinates of will give the sum as 9.

Thus, D.

Hope it helps.
Try again guys. This is a question that could really come on the GMAT. Be careful in analysis.
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by scoobydooby » Tue May 12, 2009 4:57 am
let the vertices be (a,b); (c, d) and (e, f)
the mid points are (1,2), (3,4) and (5,0).

a+c=2 (applying midpoint formula)
c+e=6
a+e=10

adding them all up, 2(a+c+e)=18
or a+c+e=9 (a, c and e are the x co-ordinates of the vertices)

hence, D

dtweah, where are we all getting mistaken? what are we missing?
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by dtweah » Tue May 12, 2009 5:20 am
scoobydooby wrote:let the vertices be (a,b); (c, d) and (e, f)
the mid points are (1,2), (3,4) and (5,0).

a+c=2 (applying midpoint formula)
c+e=6
a+e=10

adding them all up, 2(a+c+e)=18
or a+c+e=9 (a, c and e are the x co-ordinates of the vertices)

hence, D

quote]

What is the solution to the system:
x+y=2
y+x=6

I bet you wouldn't go about it by adding 2 and 6.

Why not find x and y?
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by dtweah » Tue May 12, 2009 5:36 am
scoobydooby wrote:let the vertices be (a,b); (c, d) and (e, f)
the mid points are (1,2), (3,4) and (5,0).

a+c=2 (applying midpoint formula)
c+e=6
a+e=10

adding them all up, 2(a+c+e)=18
or a+c+e=9 (a, c and e are the x co-ordinates of the vertices)

hence, D

dtweah, where are we all getting mistaken? what are we missing?
You guys are correct. D is the answer. I was thinking of another problem. My bad. I owe you all one. (laugh)
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