A right triangle with its inscribed circle. Two cevians are drawn through the circle centre. What is the circle area in terms of triangle areas A and B?

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## Solution

The circle area is 2πAB/(A+B).

First note that the two triangles both have height r, which is also the circle radius. Name their respective bases a and b. The two little triangles have bases x and y.

From triangle similarity one can show that x/r=r/(b+y) and y/r=r/(a+x). Eliminating r from these two equations leads to x/y=a/b. So we get that b+y=(a+x)b/a=bz/a, where we temporarily named a+x=z. Filling this back in we get that r^{2}=yz=xbz/a.

Now the total area is (r+z)(r+bz/a)/2. By adding up the area of the square and the four right triangles we get that this must be equal to r^{2}+rz+rzb/a. After some algebra, also using the equation for r^{2}, we get xb/a+r+rb/a-bz/a=0. Eliminating z, this leads to r=ab/(a+b).

Finally, using A=ar/2 and B=br/2, we reach the desired formula.

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## One reply on “The eye mask”

Another approach is to express A , B, and r as functions of the two side lengths, say a and b (not the usage in the Solution above), and then show that the combination 2AB/(A+B) agrees with r^2.

First, for right triangles, we have the easy formula for the radius r of the incircle: 2r = a+b – c, where c=sqrt(a^2+b^2) is the length of the hypotenuse.

Second, angle bisectors split the far side in the same ratio as the other two side lengths. The A and B triangles have heights r and bases given by that ratio property:

base_A = (c/(b+c))*a

base_B = (c/(a+c))*b

Thus

A = base_A*(r/2) = (c/(b+c))*a*(a+b-c)/4,

B = base_B*(r/2) = (c/(a+c))*b*(a+b-c)/4.

AB = c^2 a b (a+b-c)^2 / (16( b+c)(a+c))

A+B= c (a+b-c) {a/(b+c) + b/(a+c)}

= c (a+b-c) (a^2+b^2 + (a+b)c)/(4(a+c)(b+c))

4AB/(A+B) = cab (a+b-c) / (a^2+b^2 + (a+b)c)

= cab (a+b-c) / (c^2 + (a+b)c)

= ab (a+b-c) / (a+b+c)

= ab (a+b-c) ^2/ ((a+b+c)(a+b-c))

= ab (a+b-c) ^2/ ((a+b)^2-c^2)

= ab (a+b-c) ^2/ (2ab)

= (a+b-c)^2/2 = 2*r^2.