Geometry Problem. Post your solution in the comments box below.

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

Click the figure below to see the complete problem 1023.

## Saturday, June 14, 2014

### Geometry Problem 1023: Triangle, Contact, Gergonne, Product of three distances

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This problem can be solved by similiarty of quadrilaterals:

ReplyDeleteStep 1:

C1PB2Tc ~ B2PA1Ta (3 equal angles -> 2 pairs of 90 and we have BTc = BTa (Because of the tangents) ---> b2^2 = a1*c1

Notice that we have to prove that a1*b1*c1 = a2*b2*c2 /we will devide by b2^2 = a1*c1

Thus we will have to prove that b1= a2*c2/b2 ---> b1*b2= a2*c2

Step 2:

Let <CTaTb = <CTbTa = a = < TaTcTb (Because of the tangents rules and equality)

Thus < C2TbB1 = < B2TcA2 = 180 - a (as mentoned above)

Notice again that C2TbB1P ~ PB2TcA2 ( 3 equal pairs of angles)

From simple proportion we get : b1/a2 = c2/b2 Q.E.D as we had to prove in Step 1.

a1/b2=sin(PTbTa)/sin(PTaTc) due to PA1TaB2 cyclic and tangent angles

ReplyDeleteb1/c2=sin(PTaTb)/sin(PTbTa) due to PC2TbB1 cyclic and tangent angles

c1/a2=sin(PTaTc)/sin(PTaTb) due to PA2TcC1 cyclic and tangent angles

Multiply top three equations to get equation to be proved (a1b1c1)/(a2b2c2)=1

Use the lemma from https://www.facebook.com/photo.php?fbid=777498538997721&set=p.777498538997721&type=1&theater

ReplyDeleteBest regards,

S. Fulger

Problem 1023

ReplyDeleteAccording to the problem 1020 we have PC_2^2=PA_1.PB_1, PB_2^2=PA_1.PC_1, PA_2^2=PC_1.PB_1.

By multiplying by members we have PA_2^2. PB_2^2. PC_2^2=PA_1^2.PB_1^2.PC_1^2, or

PA_1. PB_1. PC_1=PA_2.PB_2.PC_2.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE