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
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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