Take any point P inside an equilateral triangle. Draw PA, PB and PC
from P perpendicular to the sides of the triangle where A, B and C
are points on the sides. Prove that PA + PB + PC is a constant.
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel lines are 1 unit and 2 units.
Prove that, given any three parallel lines, an equilateral triangle
always exists with one vertex on each of the three lines.
This solution comes from Soh Yong Sheng
(Raffles Institution, Singapore).
First he added a few lines and letters to the diagram:
Since the circles are tangent to the triangles, we know that CE
is perpendicular to GF, and AB is perpendicular to GF. [A line
touching a circle (tangent) is always perpendicular to the radius
which meets it.] We also see by symmetry that FA is perpendicular
The line FA bisects the original equilateral triangle, so we
know that FGA=60°, and GFA=30°. We can now
see that in triangle FCE, CEF=60°, and in triangle ABF,
BAF=60°. So all three of these triangles have angles
30°, 60°, 90°.
Soh Yong Sheng then used trigonometry on
these triangles. For those of you who haven't met trigonometry,
notice that a triangle with these angles is half of an equilateral
triangle, and so the shortest side is half the length of the
longest. This tells us that EF = 2CE (from triangle FCE), and FA =
2BA (from triangle ABF).
We know that CE is 1 unit, so EF = 2 units. ED is also a radius
of the small circle, so the total length of DF is 3 units.
Now in triangle ABF, FA is twice BA, but BA and DA are both radii,
and so must be the same. This means that DF is half of FA, so FA is
6 units, and DA is 3 units. This is the radius of the larger
We also received solutions from Alex and
Prateek ( Riccarton High School, Christchurch, New Zealand), and
Ramani (Australia). Well done to everyone.