Geometry Problem 1486. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

## Friday, November 27, 2020

### Geometry Problem 1486: Right Triangle, Altitude, Incircle, Tangent, Measurement

Labels:
altitude,
incircle,
inradii,
inradius,
measurement,
right triangle,
tangent

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Tr.s ABD, BDC, ABC are similar.

ReplyDeleteIf r is the inradius, r/b = 3/c = 4/a ( = 1/n say) and since a^2 + c^2 = b^2, r = 5.

So a = 4n, b = 5n and c = 3n

Now also r = ac/(a+b+c) from which 5.(4n + 5n + 3n) =12.n^2

so that n = 5 and hence a = 20, b = 25 and c = 15

So DE = CD - CE = 20^2/25 - (20 + 25 - 15)/2 = 16 - 15 = 1

Sumith Peiris

Moratuwa

Sri Lanka

From D are drown vertical two equal tangential segments

ReplyDeleteSo FP (P tg point of red circle)=DE=4-3=1

How is FP = DE?

Deletehttps://photos.app.goo.gl/9nhsrsHvmvmBh5HAA

DeleteSo centers of three circles lie on a circle with center E

Delete=> from one of righttriangles with side 3, 4 can calculate the R=5

Credits to Sumith sir for deriving InRadius = 5 and AB=15,BC=20,AC=25

ReplyDeleteBD.25=15.20

BD=12

Also 3/AD=5/15

AD=9

let P,O,Q be the centers of Green,Blue and Red incircles.M,N be the foot of perpendiculars from P and Q on AC

let AM=u, CN=v and we know DM=3, DN=4

APM similar to AOE, so AE=5u/3

CQN similar to COE, so CE=5v/4

AE+CE=(20u+15v)/12=25

4u+3v=60

But u+3+v+4=25=> u+v=18

hence u=6 & AE=10

AD=9; it follows DE=1

Another solution (using 1487):

Let X be on AD such that BX is the angle bisector of ABD

DX/XA=12/15=4/5

XA=5DX/4

AD=9DX/4=9

So DX=4

EX=Inradius=5 (O center of Blue circle. BOXA is concyclic and triangle OXE is right and isosceles)

=>DE=EX-DX=1

ABD and BCD are similar and their sides are in the ratio 3:4

ReplyDeleteLet BC=x,DC=y=>AB=3/4x,AD=9/16y

Apply Pythagoras to ABC

=>9/16x2+x2=25.25/16.16y2

=>x=5/4y

Substituting x with the above value and representing the sides of ABC in terms of y => a=5/4y, b=25/16y,c=15/16y

Semi-perimeter of ABC P=15/8y

AE=P-a=15/8y-5/4y=5/8y

Hence DE=AE-AD=5/8y-9/16y=1/16y --------(1)

To deduce the value of "y", lets first find the in-circle radius of ABC

ADB is similar to ABC => AB/3=AC/r => 15y/16.3=25y/16.r => r=5

r.P=S(ABC)

=>5.15y/8=1/2.5y/4.15y/16

=>y=16

Substituting "y" in (1) => DE=1