## Wednesday, October 21, 2015

### Geometry Problem 1155: Right Triangle, Altitude, Incenter, Incircle, Area

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below for more details. #### 5 comments:

1. Let S(XYZ) denote area of triangle XYZ
Let r1, r2, r3 are radii of incircles of triangles AHB, BHC and ABC
Note that triangles AHB, BHC and ABC are similar ( case AA)
Triangles AO1B, AOC and BO2C are similar ( case AA)
S(AO1B)/S(AOC)= (AB/AC).(AO1/AO)
Since Triangle ABH similar to ABC so we have AO1/AO= r1/r= AB/AC
So S(AO1B)/S(AOC)= (AB/AC)^2
Similarly we have S(BO2C)/S(AOC)= (BC/AC) ^2
Since AC^2=AB^2+BC^2 => S(AOC)=S(AO1B)+S(BO2C)= 13
So S(BO2C)= 13-5=8

2. Let r1 be the radius of O1 and r2 of O2 and r of. O.

Let the respective areas be S1, S2 and S

From similar Tr.s r1= rc/b and r2 = ra/b

So S1 = rc^2 / 2b and S2 = ra^ 2/2b and S = rb/2

Now since a^2 + c^2 = b^2 we have

S1(2b/r) + S2(2b/r) = S (2b/r)

Simplifying S1 + S 2 = S

So 5 + S2 = 13 and S = 8

Sumith Peiris
Moratuwa
Sri Lanka

3. See Problem 36:
Area Tr. AOC = Area Tr. AO1B + Area Tr. CO2B or 13 = 5 + Area Tr. CO2B or Area Tr. CO2B = 8 Sq. Units

4. Dear Antonio - from the data given is it possible to find the 3 sides of the Tr. ?

Rgds

Sumith

5. Easily BO1_|_CO, BO2_|_AO, and let BO1, BO2 intersect AC at M, N respectively. By symmetry tr.ANO1 and ABO1 are congruent, and so are CO2M and CO2B. Now, if MO2 and NO1 intersect at X, that is the orthocenter of tr BMN with <MBN=45, since <BO1O=(<A+ABH)/2 =45 gives <BO1N=90, and so is <BO2M, hence OO1XO2 is a parallelogram, and S[OO1XO2]=2S[O1XO2] ( 1 ).
MNO2O1 is cyclic, tr. O1XO2 and MXN are similar, their similitude ratio is O1O2/MN, but since <O1MO2=45, we get O1O2/MN=1/sqrt{2}, i.e. S[MXN]/S[O1XO2]=(MN/O1O2)^2=2; with (1) we got S[MXN]=S[OO1XO2] and indeed S[AO1B]+S[BO2C]=S[AOC], done