Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1193.

## Tuesday, March 1, 2016

### Geometry Problem 1193: 3-4-5 Right Triangle, Congruent Circles, Tangent, Radius

Labels:
3-4-5,
circle,
congruence,
radius,
special right triangle,
tangent

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Let the points of tangency on the hypotenuse be M and N. Let the parallel to BC thro O2 meet the parallel to AB thro O1 meet at L

ReplyDeleteO1M = O2M = r and they are parallel to each other hence O1O2NM is a rectangle

Let AM = p

Then MN = 2r, CN = 5-p-2r, O1L = 3-p-r and O2L = r+p-1

Tr. O1O2L is similar to Tr. ABC since corresponding sides are parallel.

So 2r/5 = (3-p-r)/3 = (r+p-1)/4

Solving the simultaneous equations for p and r

p = 10/7 and r = 5/7

Sumith Peiris

Moratuwa

Sri Lanka

What is L

Deletehttp://s22.postimg.org/3t229d6xd/pro_1193.png

ReplyDeleteLet AO1 and CO2 meet at O

Let r’ and r1 are radii of incircles of triangles ABC and O1DO2

Observe that O is the incenter of both triangles

.r’= area of Tri. ABC/half of perimeter of ABC= 6/6=1

O1DO2 and ABC are similar triangles => r1/r’= O1O2/BC= 2.r/5 => r1=2.r/5

We have r’=r1+r= 7/5.r =1 => r=5/7

Where is D pointed

Deletesee sketch below for location of D

Deletehttps://photos.app.goo.gl/dGZ6sowCZBmTynxo8

Name T1, T2 tg points of O1, O2 Draw from O1//to AB , from O2 // to BC

ReplyDeleteThey meet at P Triangle PO1O2 similar to ABC => T2C=1,5 T1A From similarity

again => AT1=2r => r=5/7