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700-800 Level Question OG 229 Geometry

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700-800 Level Question OG 229 Geometry

by EMAN » Mon Oct 05, 2009 5:37 pm
Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and y coordinates of P, Q and R are to be integers that satisfy the inequalities -4 <= x <= 5 and 6 <= y <= 16. How many different triangles with these properties could be constructed?

(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100

There has to be a relatively easy way to solve this. Please post your answer and explanation. What makes this difficult for me at least is that it involves elementary combinatorics in addition to simple coordinate geometry.
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by xcusemeplz2009 » Tue Oct 06, 2009 3:35 am
IMO C

whats the OA
It does not matter how many times you get knocked down , but how many times you get up
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Correct

by EMAN » Tue Oct 06, 2009 7:31 am
Correct. Please share your strategy.
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by xcusemeplz2009 » Tue Oct 06, 2009 9:29 am
thnks for the OA
no. of possible value for x=10 and y=11

1)information given is right angle at P, PR(ll) to x axis

information inferred y coordinates of p and r is same.

2) PQ (llel) y therefor x coordinates of P and Q are same.

now for P x can be selected in 10 ways and y in 11 ways
=10*11
for R x in 9 ways and y in 1 way( as same of P)=9*1

for q x in 1 way and y in 10 ways( one already selected for P)=10*1

tot ways=10*11*9*10=9900

HTH
It does not matter how many times you get knocked down , but how many times you get up
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