## Wednesday, October 10, 2012

### Problem 813: Right Triangle Area, Excircles, Cathetus, Tangency Points, Quadrilateral Area

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 813.

1. The area
= 30 - (1/2)(ra*rc)
= 30 - (1/2)[30/(s-a)]*[30/(s-c)]
= 30 - (1/2)*30*30/[ (1/2)(b+c-a)*(1/2)(a+b-c) ]
= 30 - (1/2)*30*30/[(1/4)*(b^2 - c^2 - a^2 + 2ac)]
= 30 - (1/2)*30*30/[(1/2)*ac]
= 30 - (1/2)*30
= 15

2. Let AB = a, BC = b, AC = c.

Denote s = 1/2 (a+b+c)
Then
BD = s−b = 1/2 (a + c − b),
BF = s−a = 1/2 (b + c − a).

S(ΔABC) = 1/2 ab
S(ΔBDF) = 1/2 (s−a)(s−b)
= 1/8 (a + c − b)(b + c − a)
= 1/8 [c^2 − (a−b)^2]
= 1/8 [c^2 − a^2 − b^2 + 2ab]
= 1/4 ab
= 1/2 S(ΔABC)

Hence, S(ADFC) = 1/2 S(ΔABC) = 15.

3. each of the above proofs airlifts in a formula with which i am completely unfamiliar. BD = s-b is a head-scratcher, at least to me.

i thought maybe the theme was to use the congruent LSs that come from the exocircles *shrugs*

note that each of the four "shortest" LSs to the unnamed tangent points is congruent to an adjacent LS that lies on the triangle of study, so:

AC + CF = AD + BD + BF
AC + AD = CF + BF + BD
given: 1/2 (AD + BD)(BF + CF) =30
and the pythagorean AC^2 = (AD + BD)^2 + (BF + CF)^2

the system of equations lacks elegance but solves much more easily than you might expect, for 1/2 (BD)(BF) = 15

1. november 1:
To Anonymous: Problem 813, Semiperimeter 's' has several theorems. See semiperimeter s theorems and problems

Thanks.

4. Some observations

2-AC = BD + BF
3- The circle centres and B are collinear