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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## Monday, May 19, 2008

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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.

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Post Comments (Atom)

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