Foot of Altitude and Minimum Distance: What is it about?
A Mathematical Droodle

 

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Explanation

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Explanation

The applet illustrates an unused problem from the International Math Olympiad submitted by France. The problem has been discussed in Crux Mathematicorum, 1987, 246, and subsequently included into R. Honsberger's collection From Erdös To Kiev.

From a point D on side BC of ΔABC perpendiculars DE and DF are drawn to AC and AB, respectively. Determine the position of D for which EF has minimum length.

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Honsberger makes an insightful observation. In case where angle at A is right the quadrilateral AEDF is a rectangle and its diagonals AD and EF are equal. So EF attains its minimum along with AD. But obviously AD has the minimum length when D is the foot of the altitude from A. He further remarks that the same holds even when angle A is not right. The reason for this is that the quadrilateral AEDF is always cyclic and, in its circumcircle, chord EF always subtends the same angle (In fact either that angle or its supplementary. But this does not affect the correctness of the claim.) It is obvious that the length of a chord subtending a fixed angle grows or decreases together with the diameter of the circle. But, since angles AED and EFD are right by the construction, AD serves as a diameter of the circumcircle. So, again, EF is the shortest whenever AD is. And we are done.

(For some reason, Honsberger makes an exception for the case where one of the base angles is obtuse. In this case, he claims, the minimum is attained for D coinciding with the obtuse angle vertex. This is incorrect.)

Gordon Walsh suggested another elegant solution. Let D' be the foot of the altitude from A and E' and F' its projections on AC and AB, respectively. Immediately

  AD' ≤ AD,

with equality only when D = D'. Observe that, since the angles with parallel sides are equal,

  ∠EDF = ∠E'D'F'

for all positions of D on BC. In a right triangle DAE,

  AD = DE / sin (DAE).

Now, since quadrilateral AEDF is cyclic so that ∠DFE = ∠DAE, we may continue applying the law of sines,

  DE / sin (DAE) = DE / sin (DFE) = EF / sin (EDF)

meaning

  AD = EF / sin (EDF) or
EF = AD sin (EDF).

This is true for all positions of D. In particular,

  E'F' = AD' sin (EDF) ≤ AD sin (EDF) = EF.

This is a nice example of a minimization problem that does not require Calculus.

References

  1. R. Honsberger, From Erdös To Kiev, MAA, 1996, pp. 43-45.

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