## Saturday, April 4, 2015

### Problem 1107: Triangle, Circumcircle, Circle, Radius, Tangent, Chord, Congruence

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

Click the diagram below to enlarge it.

1. http://s25.postimg.org/ufbka0myn/pro_1107.png

Let circle D cut AB at E and F
Since DC^2=DA.DB => Circle D is Apollonius circle of AB
And PE and CE are bisectors of angles APB and ACB
So PA/PB=EA/EB=CA/CB (1)
Triangles PBC similar to PC1B1 => B1C1= (PC1/PB). BC…..(2)
Triangles PAC similar to PC1A1=> C1A1= (PC1/PA). CA …..(3)
Divide (2) to (3) and using (1) we get B1C1= C1A1

2. Let AB cut circle D at Q.
Let < BCD = p, <BPD = q, < QPC = r and < QCP = s

Then <CAD = p and < PAD = q since DP (=DC) is a tangent to circle APB at P.
Further < PDA = 2s and < CDA = 2r so that
< PQD = < QPD = 90-s …..(1) and
< QCD = CQD = 90-r…….(2)

From (2) < ACQ = 90-p-r …..(3)
From (1) < APQ = 90-q-s ……(4)

Now < ABB1 = 2s+q & …..(5)
And < CBA = 2r+p ……(6)

Further < PCB = 90-p-r-s from which
< A1B1C1 = 90-p-q-r-s …..(7) since < PCB = < PB1C1 and < BB1A = BAA1 = q

Also < B1A1C1 = < PA1C1 - <PA1B1 = < ACC1 - < ABB1 = (90-p-r+s) – (2s+q) =
90-p-q-r-s…(8)

From (7) and (8), < A1B1C1 = < B1A1C1 and therefore C1A1 = C1B1

Sumith Peiris
Moratuwa
Sri Lanka

3. Dear Antonio Gutierrez,
I noticed that you deleted my solution in Russian to this problem. Will it be acceptable in the future
if I post side by side English-Russian translations of my solutions so that Russian readers can understand them?

1. Thanks Ivan,
Problem 1107: Could you post your solution in Russian again? I will publish it

Thanks again

2. This comment has been removed by the author.

4. Here is my solution, though rushed you should be able to deduct from my reasoning, and if necessary I may repost the solution in case the wording is difficult to understand. http://prnt.sc/eag3ym

5. Dear Barth
Referring to line 3 of your solution " from triangle A1B1C angle A1 is 90 degrees... "

Per the problem statement , P is any point on circle center D, radius DC. It may exist a particular point P so that C, O and B1 are collinear, but in general these 3 points are not collinear and angle A1 in triangle A1B1C is not equal 90 degrees. please provide explanation.

Peter Tran

6. 1. PD^2=DC^2=DB*DA, so triangles DPB and DAP are similar so <DPB=<DAP=e.
2. Circle D cuts AB at E. Call angles in triangle ABC a,b,c respectively.
<ACD=c+a, so <ADC=180-2a-c so <ECD=a+c/2 so <ECB=<ECD-<BCD=(a+c/2)-(a)=c/2.
Call <ECP=d. <ACB1=<B1BA=<PDB+<DPB=2d (since <ECP subtends arc EP of circle D)+e.
<B1CC1=<ACC1-<ACB1=(c/2+d)-(2d+e)=c/2-d-e.
<A1CC1=<BCC1-<BCA1=(<BCE-d)-(<BAA1)=(c/2-d)-(e)=c/2-d-e.
From last 2 lines, C1A1=C1B1.