## 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. 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