

the intersecting chords problem 


Next I present a geometrical problem which was published on Facebook by mr. Omid Motahed.
Three circles intersect each other at two points.
The three common cords AB, CD, EF appear to intersect at one point G.
The problem is: "why one point"?
This is not easy to prove.
However after a twist in thinking, this becomes obvious.
To demonstrate this proof I start with a seeming different problem:
"how to construct a random pyramid"?
First we define a ground plane, ABC:
Next we define the projection of the top (T) of the pyramid on the ground plane
and we draw lines through T that are perpendicular to AB, BC, AC.
Next choose the height of the piramid: say h (h perpendicular to TD), see figure below:
Points T are the same, T being located distance h above ground plane ABC.
The dotted triangle has to be rotated around line TD until h is perpendicular to plane ABC.
Take D as center of a circle with radius TD.
Next draw T_{1}A, T_{1}B.
Take B as center of circle with radius T_{1}B.
Take C as center of circle with radius T_{2}C.
Take A as center of circle with radius T_{3}A.
So, in above figure we see
BT_{1} = BT_{2}
CT_{2} = CT_{3}
AT_{1} = AT_{3}
Now rotate plane ABT_{1} around AB,
rotate BCT_{2} around BC and
rotate ACT_{3} around AC
until points T_{1} T_{2} T_{3} unite.
This is pyramid TABC.
Back to the original problem:
We paint three intersecting circles with centers A,B,C.
Now consider ABC a ground plane just as in the pyramid construction above.
Note that the red lines, the green lines and the purple lines are equal in length.
So, planes ABT_{1}, BCT_{2} and ACT_{3} may be side planes of a pyramid.
Now, rotate these planes around AB,BC and AC until the tops T meet.
Point G in the original problem is the projection of the top T on ground plane ABC.
So, the common chords intersect at one point.
Note: if plane ABT_{1} rotates around line AB, then the projection of T_{1}
on plane ABC moves over the common chord of circles A and B. Etc.
This concludes the proof.

