Proposed Problem

See complete Problem 24 at:

gogeometry.com/problem/p024_right_triangle_incircles.htm

Level: High School, SAT Prep, College geometry

## Tuesday, April 7, 2009

### Problem 24: Right triangle, Altitude, Incircles

Labels:
altitude,
incircle,
inradius,
Pythagoras,
right triangle,
similarity

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With the usual nomenclature, we've BD=ac/b and AD = c^2/b. Thus, r1 = (ca/b * c^2/b)/(ac/b+c^2/b+c) = ac^2/(b(a+b+c))

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

Therefore, r1^2 + r2^2 =

a^2c^4/(b(a+b+c))^2 + c^2a^4/(b(a+b+c))^2

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

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

= [ac/(a+b+c)]^2 = r^2

QED

Ajit: ajitathle@gmail.com

Triangle ABD similar to tri. ABC and we have r1/r = AB/AC

ReplyDeleteTriangle BDC similar to tri. ABC and we have r2/r = BC/AC

ABC is a right triangle so we have AC^2=AB^2+BC^2

And r^2=r1^2+r2^2

Peter Tran

Reference my proof for Problem 23

ReplyDeleter1 = rc/b and r2 = rc /b

So r1^2 + r2^2 = r^2(a^2 + c^2)/b^2 = r^2 from Pythagoras

Sumith Peiris

Moratuwa

Sri Lanka

https://www.facebook.com/photo.php?fbid=10206230208662809&set=a.10205987640598759.1073741831.1492805539&type=3&theater

ReplyDeleteLet the point M be the center of the circle with radius r1.

ReplyDeleteLet the point N be the center of the circle with radius r2.

Let the point O be the center of the circle with radius r.

Then draw a line from M perpendicular onto BD. The line from M meets BD at the point P.

Also draw a line from the point N perpendicular onto the radius r. This line meets

the radius at point Q.

We have to use the results from problem 23 BD=r1+r2+r and the result from problem 28 DE =r2-r1 .

PB = BD-r1 = r+r2

QN = r2-DE = r1

And

OQ = r-r2

MP = r1

Triangles BPM and NQO are similar we have :

MP/BP = OQ/NQ

r1/(r+r2) = (r-r2)/r1

r1^2 = (r+r2)(r-r2)

r1^2 = r^2-r2^2

r1^2+r2^2=r^2