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Problem 407

Geometry Difficulty 2.0 Prove it CEMC Hypatia · Canada · 2026

If points are collinear, then they lie on the same
straight line.

If the points $A(1,2)$ , $B(2,5)$ and $C(3,c)$ are collinear, what is the value of $c$ ? Three distinct points $D(0,7)$ , $E$ and $F(14,0)$ are collinear. If the coordinates of point $E$ are positive integers, how many such points are possible? Determine the value of $n$ so that the points $P(15,12)$ , $Q(6,n-4)$ and $R(18,n)$ are collinear.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Since $A$ , $B$ and $C$ are collinear, segments $AB$ and $AC$ (and $BC$ ) have the same slope. The slope of segment $AB$ is $\dfrac{5-2}{2-1} = 3$ . The slope of segment $AC$ is $\dfrac{c-2}{3-1} = \dfrac{c-2}{2}$ . Therefore, we have that $3 =
$\dfrac{c-2}{2}$ $. Solving for$ c $gives us that$ c=8 $. Since segment$ DF $has slope$ $\dfrac{7-0}{0-14}$ = - $\dfrac{1}{2}$ $and$ y $-intercept$ 7, the equation of the line through points

Figure for this problem

Figure for this problem D $and$ F $is$ $y=-\dfrac{1}{2}x$ + 7 $. The coordinates$ (x,y) $of$ E $must be positive integers and$ E must lie on the line

Figure for this problem

Figure for this problem $y=-\dfrac{1}{2}x$ + 7 $. When$ x $\geq$ 14 $we have that$ y $\leq$ 0 $, and when$ x < 14 $we have that$ y>0 $. Therefore$ x $and$ y are both positive exactly when

Figure for this problem

Figure for this problem 0<x<14 $. Also,$ y $is an integer exactly when$ x is even. Therefore the number of possible points

Figure for this problem

Figure for this problem E $is the number of integers$ 0 < x < 14 $such that$ x $is even. There are$ 6 such integers which give the following

Figure for this problem

Figure for this problem 6 $possibilities for$ E $:$ (2,6) $,$ (4,5) $,$ (6,4) $,$ (8,3) $,$ (10,2) $, and$ (12,1) $. The slope of segment$ PQ $is$ $\dfrac{n-4-12}{6-15}=\dfrac{n-16}{-9}$ $. The slope of segment$ PR $is$ $\dfrac{n-12}{18-15}=\dfrac{n-12}{3}$ $. Therefore, points$ P $,$ Q $and$ R are collinear exactly when

Figure for this problem

Figure for this problem $\dfrac{n-16}{-9}=\dfrac{n-12}{3}$ $. This is equivalent to$ 3n-48 = -9n+108 $, which is equivalent to$ 12n=156 $. Therefore, the value of$ n $that makes$ P $,$ Q $and$ R $collinear is$ $n=\dfrac{156}{12}=13\$.$

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.