Proposed Problem

See complete Problem 23 at:

gogeometry.com/problem/p023_right_triangle_incircles.htm

Level: High School, SAT Prep, College geometry

## Tuesday, April 7, 2009

### Problem 23: Right triangle, Altitude, Inradius

Labels:
altitude,
cathetus,
hypotenuse,
inradius,
right triangle

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Under the standard nomenclature, AD = c*sin(C)= c*c/b=c^2/b while BD=c*cos(C)= ac/b ----(1)

ReplyDeleteSo r1 = AD*BD/(AD+BD+AB)

= (c^2/b*ac/b)/(c^2/b + ac/b + c)

= ac^2/b(a+b+c)

Similarly, r2 = ca^2/b(a+b+c)

and r = ac/(a+b+c)

So,r1+r2+r=ac^2/b(a+b+c)+ca^2/b(a+b+c)+ac/(a+b+c)

= ac/b = BD by equation (1)

Ajit: ajitathle@gmail.com

The statement that r=ac/(a+b+c) is incorrect; the numerator should be 2ac.

ReplyDeleteHowever, it is easier to show that r=a+b-c and use that to obtain the desired result.

Prof Elie Achkar (Lebanon)

ReplyDeleteBD + DC = BC + 2r1 in the triangle BDC

AD + BD = AB + 2r2 in the triangle BDA

AB + BC = AC + 2r in the triangle ABC

then :

BD+DC+AD+BD+AB+BC = 2r1 + 2r2 + 2r

THEN :

2BD = 2(r1 + r2 +r )

follow : BD = r1 + r2 + r

Inradius = Tr. Area/Semisum = (ac/2)/[(a+b+c)/2]Hence r = ac/(a+b+c)

ReplyDeleteSo then why is the statement incorrect?

Dear Joe = Ajit:

ReplyDeleteI agree, your statement r = ac/(a+b+c) is correct.

How is it possible that

ReplyDeleteBD + DC =BC + 2r1

.................

.................

Please tell me the condition no

In a right triangle ABC (right angled at B)

ReplyDeleter=s–b,is well-known.

Easily remembered as :

in-radius = semi perimeter–hypotenuse

(or)

in-diameter=perimeter–twice the hypotenuse

So 2(r + r1 + r₂)

=(AB+BC–AC)+(BD+AD–AB)+(BD+DC–BC)

=2BD+(AD+DC)–AC

=2BD+AC–AC

=2BD

Hence r+r1+r₂=BD

If O1, O2 and O are the centers of circles with radius r1,r2 and r then

ReplyDeleter1 = rc/b and r2 = ra/b since Tr.s AO1C, CO2B and AOC are similar

So r+r1+r2 = r (a+b+c)/b ....(1)

By writing the area of the right Tr. in 2 ways

BD. b = r(a+b+c)

So from (1) r+r1+r2 = BD

Sumith Peiris

Moratuwa

Sri Lanka