Waste less time on Facebook — follow Brilliant.


\(ABC\) any triangle

Any \(BD\) and \(CE\) drawn such that \(D\) lies on \(AC\) and \(E\) lies on \(AB\)

\(DE\) drawn

\(BD\) and \(CE\) intersect at \(I\)




Prove that \(M-N-O\) are collinear

Note by Megh Parikh
4 years, 1 month ago

No vote yet
9 votes


Sort by:

Top Newest

We will use normalized barycentric coordinates \(A = (1,0,0)\), \(B = (0,1,0)\), \(C = (0,0,1)\). Then \(D = (p,0,1 - p)\), \(E = (q,1 - q,0)\). Now \(N\) is a midpoint of \(DE\), hence \[N = \left(\frac{p + q}{2}, \frac{1 - q}{2},\frac{1 - p}{2}\right)\] Equation of line \(BD\) is \[pz = (1 - p)x\] and equation of line \(CE\) is \[(1 - q)x = qy\] Intersecting these lines we get \[I = \left(\frac{pq}{p + q - pq}, \frac{p(1 - q)}{p + q - pq}, \frac{q(1 - p)}{p + q - pq}\right)\] Now \(O\) is midpoint of \(AI\), hence \[O = \left(\frac{p + q}{2(p + q - pq)}, \frac{p(1 - q)}{2(p + q - pq)}, \frac{q(1 - p)}{2(p + q - pq)}\right)\] And \(M\) is midpoint of \(BC\), so trivially \[M = \left(0, \frac{1}{2}, \frac{1}{2}\right)\] Now note that \[\begin{vmatrix} \frac{p + q}{2}& \frac{1 - q}{2}&\frac{1 - p}{2} \\ \frac{p + q}{2(p + q - pq)}& \frac{p(1 - q)}{2(p + q - pq)}& \frac{q(1 - p)}{2(p + q - pq)} \\ 0& 1/2& 1/2 \end{vmatrix} = 0\] Hence \(M,N,O\) are collinear.

Jan J. - 4 years, 1 month ago

Log in to reply

I'm not sure I understand your last step. Are you taking the determinant of a matrix? I'm assuming that's what you're doing. How does that tell you that the three points are co-linear?

Sotiri Komissopoulos - 4 years, 1 month ago

Log in to reply

The main idea is: If the area of a triangle is 0, then the vertices lie on a straight line.

Note that once you found the coordinates of the points, you could go about finding the line that these three points are on. However, unless the ratios are immediate, this often requires a bunch more algebra. Showing that the area is 0 could be more straightforward.

Barycentric coordinates merely make the algebra much easier / nicer. A coordinate geometry approach would work similarly, though with much uglier values.

Calvin Lin Staff - 4 years, 1 month ago

Log in to reply

See here http://zacharyabel.com/papers/Barycentric_A07.pdf page 5.

Jan J. - 4 years, 1 month ago

Log in to reply

This is the Newton-Gauss line of a complete quadrilateral.

Jon Haussmann - 4 years, 1 month ago

Log in to reply

Why did every comment gets vote down?

Zi Song Yeoh - 3 years, 10 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...