Let ABC be a 3-4-5 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?
|CD|^2 = |AC|^2 + |AD|^2 - 2|AC||AD|cos(A)
y^2 = 4^2 + x^2 - 2*4*x*(4/5)
= x^2-32*x/5+16 (1)
The inradius of a triangle is equal to the
area divided by the semiperimeter.
2*Area(ACD) 2*Area(BCD)
---------------- = ----------------
Perimeter(ACD) Perimeter(BCD)
|AC||AD|sin(A) |BC||BD|sin(B)
----------------- = -----------------
|AC|+|CD|+|AD| |BC|+|CD|+|BD|
4*x*(3/5) 3*(5-x)*(4/5)
----------- = ---------------
4+y+x 3+y+(5-x)
20-7*x
y = -------- (2)
2*x-5
Combining (1) and (2) give
(20-7*x)^2
------------ = x^2-32*x/5+16
(2*x-5)^2
or
x*(5*x^3-57*x^2+210*x-250) = 0
or
x*(x-5)*[5*x-16+-sqrt(6)] = 0
Therefore,
x = 0, [16-sqrt(6)]/5, [16+sqrt(6)]/5, or 5.
We can disregard 0 and 5. Plugging
[16+sqrt(6)]/5 into (2) gives y < 0;
thus we can eliminate it also. Plugging
[16-sqrt(6)]/5 into (2) gives y = sqrt(6);
which is confirmed by Geometer's SketchPad.
Therefore,
|AD| = x = [16-sqrt(6)]/5 and |CD| = y = sqrt(6),
where A is the vertex of triangle ABC opposite
the leg of length 3.
QED
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Posted by Bractals
on 2017-01-01 11:00:32 |
No Subject
