perplexus.info :: Calculus : Minimum perpendicular

Minimum perpendicular (Posted on 2007-05-09)

Let X be a point in the interior of triangle PQR.
Let a line through X intersect rays QP and QR in points A and B respectively.
Let Y be the point on line segment AB such that |BY| = |AX|.

Prove that |AB| is a minimum if and only if AB is perpendicular to QY.

Submitted by Bractals

Rating: 2.6667 (3 votes)

Solution: (Hide)

Let a = |AX| = |BY|, b = |BX| = |AY|, p = |QX|,
u = measure of angle AQX, v = measure of angle BQX,
x = measure of angle QAB, and y = measure of angle QBA.

From triangles AQB, AQX, and BQX we get
d = a + b (1) u + v + x + y = 180 (2) a sin(x) = p sin(u) (3) b sin(y) = p sin(v) (4)
Let z' denote dz/dx. Taking the derivative of these four equations gives
d' = a' + b' (1') 1 + y' = 0 (2') a cos(x) + a' sin(x) = 0 (3') b cos(y) y' + b' sin(y) = 0 (4')
To minimize d we set d' to zero. Eliminating the primed terms from the primed equations gives
b tan(x) = a tan(y) or |AY| tan(QAY) = |BY| tan(QBY)
Which is equivalent to AB being perpendicular to QY.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re(3): minimum?	Charlie	2007-05-10 12:29:17
re(2): minimum?	Dej Mar	2007-05-10 11:27:35
re: minimum?	JayDeeKay	2007-05-10 09:03:18
minimum?	Dej Mar	2007-05-10 07:40:57

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