## Monday, July 23, 2012

### Problem 789: Right Triangle, Altitude, Angle Bisector, Incircle, Tangency Point, Inradius, Circles, Tangent Line

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

Click the figure below to see the complete problem 789.

1. http://img138.imageshack.us/img138/1382/problem789.png

( complete solution)
drawing lines and circle per attached sketch
We have Arc B’F= Arc FC and Arc AB=Arc AB’
∠ABD=1/2(Arc AB’ +Arc B’F)
∠ADB=1/2(Arc AB +Arc FC) = ∠ABD
So ABD is an isosceles triangle with AB=AD
We also have ∠HBD=∠DBK=∠BDK.. ( alternate angles)
So BKD is an isosceles triangle with KB=KD
A,I and K are collinear ....(located on perpen. Bisector of BD)
IK is an angle bisector of ∠ BKD
So IM=IL=ED=r

1. Anonymous said...

AB=AD so we put M and N the match points with AB and BC .

ED=r

July 23, 2012 7:35 PM

2. CD/HD = BC/BH or CD = CH *BC/(BC+BH) = a^2/(b+c). Now let the in-circle touch AB at F. AE =AF = c - r or CE = b-c+r which gives us:DE = CE-CD=b-c+r-a^2/(b+c)= (b^2-c^2-a^2)/(b+c) + r = r since b^2=a^2+c^2. Thus, DE = r

3. Simply let AB = a, BC = b, AC = c

By using the properties of inradius, r = ab/(a+b+c)
Our target is DE = ab/(a+b+c)

The altittude = ab/c
CH = sqrt(b^2-(ab/c)^2) = (b^2)/c
Using angle bisector theorem, CD = CH*(BC/(BC+CH)) = (b^2)/(a+c)
By incircle properties, CE = (b+c-a)/2
Then DE = CE - CD = (b+c-a)/2 - (b^2)/(a+c)
By trigonometry, we can easily prove the remainings.

4. Anonymous

1. AB=AD We put M and N the points where they match with AB and BC

IM=EN=NB=BM=r
AM=AE AD-AE= AB-AM=r so in this case ED=r

5. http://img507.imageshack.us/img507/9789/p781resuelto.png
(look at the picture)

Let be P the intersection point of the ray AI and segment BC. Note that as triangle ABC is rectangle, we know that <BAC=<HBC, but AI and BD are bisectors, then <DAP=<DBP, i.e, the quadrilateral ADPB is cyclic. immediately <PDA=90° and BP=DP. As triangle DBP is isosceles, the incircle is tangent to DP too.
Now let Q be the tangency point of incircle with DP. Note that IQED is a square of side r. Therefore DE=r

Greetings :)

1. Look at THIS picture! Sorry :P

http://img27.imageshack.us/img27/8784/p789resuelto.png

6. Let F be the tangency point on AB

< IBC = 45 and < DBC = 45-C/2 so < IBD = C/2

Hence BCDI is cyclic and BI = ID since IC bisects < C

So Tr.s BFI and DEI are congruent (SS & right angle)

Therefore DE = BF = r

Sumith Peiris
Moratuwa
Sri Lanka

7. Or because BCDI is cyclic and < IDE = IBC = 45 the result follows

8. Problem 789
Let <ABH=<BCA=x and <HBD=<DBC=y then <ABD=x+y=<ADB so AB=AD.Si the incircle of radius r is tangent to AB at F then AF=AE,BF=ED.But <IBF=45 so ED=FB=IF=r.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE