Asian Pacific Kite

A problem from the 1993 Asian Pacific Mathematical Olympiad led Vo Duc Dien to the following extension:

Let ABCD be a quadrilateral such that all sides have equal length and angle ABC is 60°. Let l be a line passing through D and not intersecting the quadrilateral (except at D). Let E and F be the points of intersection of l with AB and BC respectively. Let M be the point of intersection of CE and AF. Find the locus of point M.

The applet below illustrates the configuration.

 

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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Discussion

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

Let ABCD be a quadrilateral such that all sides have equal length and angle ABC is 60°. Let l be a line passing through D and not intersecting the quadrilateral (except at D). Let E and F be the points of intersection of l with AB and BC respectively. Let M be the point of intersection of CE and AF. Find the locus of point M.

In fact, the configuration exhibits one additional property. Let C(AEM) and C(CFM) be the circumcircles of the respective triangles. The two circles intersect in M and one other point N. Is can be seen in the applet, N is collinear with EF.

 

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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

As was originally shown, CA² = CM × CE. Similarly, CA² = AM × AF. By the intersecting secants theorem, this means, in particular, that the circles C(AEM) and C(CFM) are tangent to AC.

It was established that ∠AMC = 120°, for the configuration where E and F lie exterior to ΔABC. Thus, for this configuration, M lies on the arc AC of the circumcircle C(ABC). As the applet illustrates, for other positions of E and F, M traces the whole of that circle. (The proof must of course be modified, for, when M is above line AC, ∠AMC = 60°.)

Now, let's show that N is collinear with E and F.

Since in C(AEM), inscribed ∠AME = 60° and is subtended by arc AE, arc(AE) = 120°. In C(CFM), ∠CMF = 60° and arc(CF) = 120°. Denoting, as before, α = ∠AEC = ∠CAF and β = ∠ACE = ∠AFC, α + β = 60°.

Consider angles MNE and MNF. ∠MNE is inscribed in C(AEM) and is subtended by the arc(EAM) = arc(AE) + arc(AM). On the other hand, ∠ACE = (arc(AE) - arc(AM))/2 so that arc(AM) = 120° - 2β, implying

  ∠ANM = 60° - β so that
∠MNE = ∠ANE + ∠ANM = 120° - β.

Similarly, ∠MNF = 120° - α. It follows that

  ∠MNE + ∠MNF = 240° - (α + β) = 240° - 60° = 180°.

Which shows that ∠ENF is straight and the three points are indeed collinear.

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62315359

Search by google: