Geometry Problem. Post your solution in the comments box below.

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

Click the figure below to view more details of problem 1171.

## Thursday, December 10, 2015

### Geometry Problem 1171: Area of a Triangle in terms of the three altitudes or heights

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I use subscript 1 2 3 instead.

ReplyDeleteSince K = 1/2 ah₁ = 1/2 bh₂ = 1/2 ch₃

a = 2K/h₁

b = 2K/h₂

c = 2K/h₃

Substitude into the result of the last problem, done.

Let p, q, r be the reciprocals of the altitudes drawn from A, B, C respectively.

ReplyDeleteSo a = 2Kp, b = 2Kq and c = 2Kr

Per the result of Problem 1170 and my proof therein

16K^2,= 2a^2b^2 + 2b^2c^2 + 2c^2a^2 - a^4- b^4 - c^4 = (a+b+c)(a+b-c)(a-b+c)(-a+b+c) where I have used the difference between 2 squares to simplify the algebra

This is = to 16K^4 (p+q+r)(p+q-r)(p-q+r)((-p+q+r) from which the result follows by dividing both sides of the equation by 16K^4 and taking the reciprocal and then the square root of both sides

Sumith Peiris

Moratuwa

Sri Lanka

calculate K.(1/ha+1/hb+1/hc)= 1/2(a+b+c)= p

ReplyDeletesimilarly K(-1/ha+1/hb+1/hc)= 1/2(-a+b+c)=p-a

....replace these value to the right hand side of the expression, we will get the result