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The figure shows rectangle $ABCD$ with segment $PQ$ dividing the rectangle into two congruent squares. How many right triangles can be drawn using three of the points $\{A,P,B,C,Q,D\}$ as vertices

 Apr 1, 2020
edited by helpppp  Apr 1, 2020
edited by helpppp  Apr 1, 2020
edited by helpppp  Apr 1, 2020
 #1
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Here is a cheaty way to do it:

 

You can count the total number of ways 3 points get selected out of the 6

\({6 \choose 3}=20\) (Order does not matter)

 

Now, there are some cases that cannot be counted (straight lines)

DQC and APB are straight therefore are not triangles.

 

So 20 - 2 = 18? Counting and probability are my weakness I'm not sure if this is right.

 Apr 1, 2020

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