### Golden Thoughts

Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.

### From All Corners

Straight lines are drawn from each corner of a square to the mid points of the opposite sides. Express the area of the octagon that is formed at the centre as a fraction of the area of the square.

### Star Gazing

Find the ratio of the outer shaded area to the inner area for a six pointed star and an eight pointed star.

# Triangle in a Triangle

##### Stage: 4 Challenge Level:

On the diagram the points that divide each of the sides into equal thirds can be marked. The lines connecting these points to the nearby vertex of the yellow triangle can also be drawn. This gives the following diagram:

In this diagram, $AE=EG=GC$, $AH=HF=FB$ and $CD=DI=IB$, since the points trisect the sides.

Now, triangle $AHE$ is an enlargement of $ABC$ by scale factor $\tfrac{1}{3}$, as $AE=\tfrac{1}{3}AC$, $AH=\tfrac{1}{3}AB$ and $\angle EAH = \angle CAB$.

This means the area of $EAH$ is $\left( \tfrac{1}{3} \right)^2 = \tfrac{1}{9}$ of the area of $ABC$.

Since $HF = AH$, $EAH$ and $EHF$ have the same base length and the same perpendicular height (that of $E$ above $AB$), they have the same area: $\tfrac {1}{9}$ of the total area of $ABC$.

This process can be repeated at vertices $B$ and $C$, so each of the six orange triangles all have area $\tfrac{1}{9}$ of that of $ABC$.

Therefore the yellow area is $1-6\times \tfrac{1}{9} = \tfrac{1}{3}$ of the area of the whole triangle.

Steven, from Sunderland College, sent us a solution which used the sine rule for area instead. You can see his solution here.