### At a Glance

The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

### No Right Angle Here

Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.

### A Sameness Surely

Triangle ABC has a right angle at C. ACRS and CBPQ are squares. ST and PU are perpendicular to AB produced. Show that ST + PU = AB

# Terminology

##### Stage: 4 Challenge Level:

Correct solutions from Mary of Birchwood High School and Andrei of School 205 Bucharest have contributed to the edited solution below.

As triangle DEF is equilateral, all its angles are $60$ .

Angle AEF $= 180 - 60 - c = 120 - c$

Similarly

Angle BFD $= 120 - b$
Angle EDC $= 120 - a$

From triangle FAE, I calculate angle A:

As triangle ABC is an isosceles Angle ABC $=$ Angle ACB $= x$

Therefore $180 - (120 - b + a ) = 180 - (120 - a + c )$
Therefore $60 + b - a = 60 + a - c$
Therefore $2a = b + c$
Therefore $a = (b + c)/2$

Part Two

If $a=b=c$

Angle ABC $= 180 - (120 - b + a ) = 180 - (120 - a + a ) = 60$
Angle ACB $= 180 - (120 - a + c ) = 180 - (120 - a + a ) = 60$
Therefore angle BAC is $60$

Therefore triangle ABC is equilateral.