## Thursday, November 17, 2022

### Geometry Problem 1502: Right Triangle, Incircle, Inradius, Geometric Mean of 2 Inradii, Angle Bisector, Perpendicular, Tangential Quadrilateral

Geometry Problem 1502. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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1. If the inradius of Tr. ABC = r and the semi perimeter = s,
Then r = S(ABC)/s = (ac/2) / [(a+b+c)/2) = (a+c - b)/2 = s-b ....(1)

Now A,E,D are collinear so from similar Tr.s, e/r = (s-a-e)/(s-a) = (s-a)/(r+s-a)
So e = (s-b)(s-a)/c ....(2) since r+s-a = c where we have used r = s-b from (1)
Similarly f = (s-b)(s-c)/a ....(3)

(2) X (3) gives e.f = (s-b)^2 . (s-a)(s-c)/ac
e.f = (s-b)^2 (b+c-a)(b+a-c)/(4ac) = (s-b)^2. {b^2 - (c-a)^2}/(4ac) = (s-b)^2 / 2
(since a^2 + c^2 = b^2)

Now BD^2 = 2.r^2 = 2(s-b)^2 from (1)

Hence BD^2 = 4ef and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

2. https://photos.app.goo.gl/46pkyTFQ4edo5UjX8

let a, b, c are sides of triangle ABC with perimeter 2p and inradius r
let DG and DM cut AC at P and Q ( see sketch)
Note that f and e are inradius of triangles QMC and AGP
and triangles QMC and AGP are similar to triangle ABC
Note that ration of inradius of 2 similar triangles will be ratio of 2 triangles

We have CM= p-c= (a+b-c)/2 and AG=p-a = (-a+b+c)/2
so f/r= CM/CB= (a+b-c)/2a
and e/r=AG/AB= (-a+b+c)/2c
so (e.f)/(r.r)= (-a+b+c) x( a+b-c)/(2a.2c)= (b^2-(a-c)^2)/(4ac)....... (1)
replace b^2=a^2+c^2 in (1) and simplify we get
(e.f)/(r.r)= 1/2 and the result follow