OG Problem Solving # 248

[Guest]

Hi all,

I want to clarify something in the answer explanation for Problem Solving #248 on p. 270 of the OG. Here is the question and answer explanation:

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

Geometry + Arithmetic Simple coordinate geometry + Elementary combinatorics

In the xy-plane, right triangle PQR is located in the rectangular area determined by -4<x<5 and 6<y<16. (There is an illustration in the book.)

Since the coordinates of points P, Q, and R are integers, there are 10 possible x values and 11 possible y values, so point P can be any one of 10(11) = 110 points in the rectangular area.

Since R has to be horizontal with respect to P, it has the same y value as P and can have 9 other x values. Q has to be vertical with respect to R, so it has the same x value as R and can have 10 other y values. This gives 110(9)(10) = 9,900 possible triangles.

The correct answer is C.

When the answer explanation says: "Q has to be vertical with respect to R, so it has the same x value as R and can have 10 other y values." Is R a typo, since the right angle is at P so Q should be vertical with respect to P? Please verify. Thank you.

[StaceyKoprince]

Excellent! That is indeed a typo - Q should be located vertical to P, not R.