Let ABC be a 345 triangle with right angle C. Let D be a point on the hypotenuse. CD then partitions ABC into ACD and BCD.
Where is D located if ACD and BCD have congruent incircles?
An elementary, but involved, construction.
Note first, that for the construction to be valid, the centres of the two circles must have the same height (one radius) from the hypotenuse, and must be the same distance from the other original enclosing side.
After constructing the 345 triangle (AB=3, AC=4), we can start by bisecting the angles at A and B. The circles are congruent and of the required size when their centres are at distances of 2 (=4^(1/2)) and 3^(1/2) from C. As it's easier to use an integer, draw circle c with radius 2 on C, and mark D, its intersection with the angle bisector of B. This is the centre of the first congruent circle. A line through D parallel to AB crosses the other angle bisector at E. This is the centre of the second congruent circle. Each can be completed by dropping a perpendicular to AB at F,G respectively and making DF and EG the radii of the circles. (It's not essential to do this, as the construction only needs the midpoint of DE, namely G.)
By construction, a common tangent (the others can be ignored) of the circles must fall on the line CG, which can be produced to AB to determine H, the point described as D in the problem.
We might not want to do all this construction every time we draw the figure, so find the midpoint of CH at J, and draw the circle d on J with radius CJ. This intersects circle C at K and L. CK and CL are known to be of length 2. HK and HL are of length 2^(1/2). As CH is a diameter, and K and L are on the circumference of circle d, CH has a length of 6^(1/2).
So the next time we draw the figure, the required point H (or D in the problem) is that one of the two points on the circle radius sqrt(6) from C, that crosses line AB closer to B.
A very nice problem.

Posted by broll
on 20170101 11:33:55 