Page 26 - MATINF Nr.2

26 T. Kalogerakis

Applications

Problem 1. Consider an A-righted triangle
ABC. From the midpoint P of arc BC (not
containing A) of its circumcircle, we draw per-
0
0
pendiculars PB 1 = b , PC 1 = c to AC = b,
AB = c respectively.
b c
Prove: + = 2.
b 0 c 0

Figure 2
0
Solution: Since a = BC = 2R = 2 · PA 1 = 2a , identity (1) gives directly the desired result.
Problem 2. Consider an equilateral triangle
ABC. From a point P of arc BC (not contai-
ning A) of its circumcircle, we draw perpen-
0
0
diculars PA 1 = a , PB 1 = b and PC 1 = c 0
to BC, AC and AB respectively.
1 1 1
Prove: = + .
a 0 b 0 c 0

Figure 3
Solution: Since a = b = c, identity (1) gives directly the desired result.

Problem 3. Consider a special triangle
ABC with b + c = 2a. From the midpoint
P of arc BC (not containing A) of its cir-
0
cumcircle, we draw perpendiculars PA 1 = a ,
0
0
PB 1 = b and PC 1 = c to BC, AC and AB
respectively.
Prove: A ABC = 3 · A PBC .

Figure 4
0
0
0
Solution: Obviously b = c = d, and identity (1), using b + c = 2a, gives a = d/2 and
A PBC = ad/4.
But, A ABC = A ABPC −A BPC = (cd/2+bd/2)−(ad/4) = (2ad/2)−(ad/4) = 3ad/4 = 3A PBC ,
so A ABC = 3 · A PBC .