# Prove that the sum of the areas of triangles $FAK$, $KCB$, and $CFL$ is equal to half of the hexagon.

Let $$ABCDEF$$ be a convex hexagon with $$\angle A=\angle D$$ and $$\angle B=\angle E$$. Let $$K$$ and $$L$$ be the midpoints of the sides $$AB$$ and $$DE$$ respectively.

Prove that the sum of the areas of triangles $$FAK$$, $$KCB$$, and $$CFL$$ is equal to half of the area of the hexagon if and only if $$\frac{BC}{CD}=\frac{EF}{FA}$$

I cant figure out how to relate the proportions of the sides to the area, suggestions as-well as solutions would be appreciated

Taken from the 2013 Pan African Maths Olympiad http://pamo-official.org/problemes/PAMO_2013_Problems_En.pdf

• @dfnu yes sorry – Tyrone Nov 19 '19 at 13:22

Consider quadrilateral $$ABCF$$. There is an elegant formula for the areas:-

$$\text {Area }CFK=\text {Area }AKF+\text {Area }BCK- \frac{1}{2}AF.BC.\text{sin}(A+B).$$

Proof

Drop perpendiculars from $$C$$ and $$F$$ onto the line $$AB$$ and let $$AK=KB=K$$. Then

$$2\times\text {Area }AKF=kAF\text{sin}A,\,\,\,\,2\times\text {Area }BKC=kBC\text{sin}B$$.

$$2\times\text {Area }FKC=(2k-AF\text{cos}A-BC\text{cos}B)(AF\text{sin}A+BC\text{sin}B)$$

$$-(k-AF\text{cos}A)AF\text{sin}A-(k-BC\text{cos}B)BC\text{sin}B.$$ Multiplying out now gives the required result.

Similarly, for quadrilateral $$DEFC$$:-

$$\text {Area }FCL=\text {Area }CDL+\text {Area }EFL- \frac{1}{2}DC.EF.\text{sin}(D+E).$$

For equal areas we require $$\text {Area }FCL+\text {Area }AKF+\text {Area }BCK=\text {Area }CFK+\text {Area }CDL+\text {Area }EFL$$ and therefore

$$\frac{1}{2}AF.BC.\text{sin}(A+B)=\frac{1}{2}DC.EF.\text{sin}(D+E).$$ i.e. if and only if $$AF.BC=DC.EF$$. (Or $$A+B=D+E=\pi$$?)

• How did you get the $\frac{1}{2}AF.BC.sin(A+B)$. Wouldn’t that only work if $AB$ was parallel to $FC$? – Tyrone Nov 20 '19 at 12:35
• If you work out the areas of the three triangles in terms of the lengths and angles, AK,AF,A etc then you'll see how it all drops out. I will return to the website later today if you have any further questions. – S. Dolan Nov 20 '19 at 12:50
• I've written out the formulae now. There was a typo (a wrong sign) which may have worried you - sorry if that was the case. – S. Dolan Nov 20 '19 at 18:20