## Sunday, June 5, 2016

### Geometry Problem 1224: Triangle, Parallel to the Sides

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

Click the figure below to view more details of problem 1224.

1. Problem 1224
Is BA2/BC=AB2/AC=AC2/AB=BC1/AB, then AC2=BC1.But AC2/C2C1=AB2/B2B1=BA2/A2A1=BC1/C1C2 .Therefore C1A2//C2A1//AC.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS PIRAEUS GREECE

2. Let m= CA1/CB
We have CB1/CA= CA1/CB= BC1/BA=BA2/BC= m
So BA2=CA1 and BA1/BC= 1-m ….(1)
We also have CA2/CB=1-m= CB2/CA=BC2/BA … (2)
Compare (1) and (2) we have BC2/BA= BA1/BC = > C2A1// AC//C1A2

3. CB1C1A2 and AB2A2C1 are parallelograms

Hence AB2C2, A1B1C and A2BC1 are all congruent triangles.

So BC1 = A1B1 = AC2

Hence AB1A1C2 is a parallelogram

So C2A1//C1A2

Sumith Peiris
Moratuwa
Sri Lanka

4. We use Barycentric Coordinates.
Definition: Capital letters will denote vectors. P = x A + y B + z C will be written as P = (x, y, z). By definition, x + y + z = 1. AB means “length of AB” unless otherwise specified; most importantly, it does NOT mean “A times B”.

If line AB is divided by C in ratio m : n, then [n / (m+n)]A + [m / (m+n)]B = C
Let A1 = (0, p, (1-p)), A2 = (0, q, (1-q)).

Because line (A1)(B1) is parallel to line AB, B(A1) : (A1)C = A(B1) : (B1)C. This indicates that B1 = (p, 0, (1-p)).
Similarly, B2 = (q, 0, (1-q)), C1 = (p, (1-p), 0) C2 = (q, (1-q), 0).
It’s given that line (C1)(A2) is parallel to line AC. Hence, 1 – p = q.
Thus, C2 = ((1-p), p, 0). In comparison, A1 = (0, p, (1-p)).
The coordinates for C indicate that A(C2) : (C2)B = B(A1) : (A1)C. Hence, line (C2)(A1) is parallel to line AC. Line AC is also parallel to line (C1)(A2), hence, line (C2)(A1) is parallel to line (C1)(A2).