Geometry Problem

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

Click the figure below to see the complete problem 813.

## Wednesday, October 10, 2012

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

Labels:
area,
excircle,
quadrilateral,
right triangle,
tangency point

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The area

ReplyDelete= 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

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

ReplyDeleteDenote 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.

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.

ReplyDeletei 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

november 1:

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

Thanks.

Some observations

ReplyDelete1-AD = CF

2-AC = BD + BF

3- The circle centres and B are collinear