See complete Problem 36
Right triangle, altitude, incircles. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 36
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See complete Problem 36
Right triangle, altitude, incircles. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
We've BD = ca/b, AD = c^2/b and therefore,
ReplyDeleter1=(ac/b+c^2/b)/(c+ca/b+c^2/b)= ac^2/(b(a+b+c))
Likewise, r2 = ca^2/(b(a+b+c)) and r=ac/(a+b+c)
Hence, LHS = ac^3/(b(a+b+c))+ca^3/(b(a+b+c))
= ac(c^2+a^2)/(b(a+b+c))
= acb/(a+b+c) = r*b since r=ac/(a+b+c)
Thus, r1*c + r2*a = r*b or r1*AB+r2*BC=r*AC
QED
Ajit: ajitathle@gmail.com
As ADB, BDC, and ABC are similar, AB/r1=BC/r2=AC/r=k, so AB=kr1, BC=kr2, AC=kr.
ReplyDeleteFrom AB^2+BC^2=AC^2 we get AB.kr1+BC.kr2=AC.kr and finaly r1.AB+r2.BC=r.AC.
Our proof is based on the simple fact that the in-radius of a triangle XYZ right angled at Z is given by (XZ+YZ-XY)/2.
ReplyDeleteAccordingly,
r1 = (AD + DB - AB)/2 and r = (AB + BC - AC)/2
So r1/r = (AD + DB - AB)/(AB + BC - AC)
= (AB/AC).[AD/AB + DB/AB - 1]/[AB/AC + BC/AC - 1]
= cos A.(cos A + sin A - 1}/(cos A + sin A - 1}
= cos A
Similarly r2/r = cos C
Hence
r1.AB + r2.BC
= r1.c + r2.a
=(r cos A)c + (r cos C)a
= r(a cos C + c cos A)= r.b
= r.AC
By similar triangles ADB and ABC we have
ReplyDeleter1/AD=r/AB
(1.) r1AB=rAD
By similar triangles BDC and ABC we have
r2/DC=r/BC
(2.)r2BC=rDC
Adding equations 1 and 2 and using AD+DC=AC
r1AB+r2BC=r(AD+DC)
r1AB+r2BC=rAC
If two triangles are similar, then the proportion is the same between their sides and their incircles (2r=AB+BC-AC)
ReplyDeleteTriangle ADB is similar to triangle ABC (angBAC, 90)
therefore AB/AC=r1/r , AB^2= AB.AC.r1/r
Triangle CDB is similar to triangle CBA (angACB, 90)
therefore BC/AC=r2/r, BC^2=BC.AC.r2/r
Triangle ABC is right in B : AC^2=AB^2+BC^2 :
AC^2= AB.AC.r1/r + BC.AC.r2/r
AC^2= (AB.r1 + BC.r2 )AC/r
therefore AC.r= AB.r1 + BC.r2
Let the incentres be I1, I2 and I
ReplyDeleteTr.s AI1B, CI2B and AIC are similar
So r1/c = r2/a = r/b = r1c/c^2 = r2a/a^2 = (r1c + r2a)/(c^2 + a^2) = (r1c + r2a)/b^2
Hence rb = r1c + r2a
Sumith Peiris
Moratuwa
Sri Lanka
I don’t understand how the addition r1/c+r2/a result in (r1c + r2a)/(c2+a2)
DeleteSimple algebraic identity : If x/y = u/v then each each fraction = (x+u)/(y+v)
DeleteSo r/b = r1c/c^2 = r2a/a^2 = (r1c+r2a)/(c^2+a^2) = (r1c+r2a)/b^2
whence rb = r1c + r2a