# Sunday Times Teaser 2545 – Emblematic

### by H Bradley and C Higgins

#### Published: 3 July 2011 (link)

Pat’s latest art installation consists of a large triangle with a red border and, within it, a triangle with a green border. To construct the green triangle, he drew three lines from the vertices of the red triangle to points one third of the way (clockwise) along their respective opposite red sides. Parts of these lines formed the sides of the green triangle. In square centimetres, the area of the red triangle is a three-digit number and the area of the green triangle is the product of those three digits.

What is the area of the red triangle?

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The diagram shows the red and green triangles and the $$(x, y)$$ coordinates of the points $$A$$, $$B$$, $$C$$, $$D$$, $$E$$ and $$F$$. Point B is at position $$(b, h)$$, point C at $$(c, 0)$$ and the points D, E and F divide the three sides in the ratios of $$r:s$$ with $$r + s = 1$$ (in our case $$r=1/3$$ and $$s=2/3$$).
We can find the coordinates of point G from the intersection of the vectors CD and AE:$(c,0)+k(rb-c,rh)=\overline{k}(sb+rc, sh)$where $$k$$ and $$\overline{k}$$ are determined by equating the $$x$$ and $$y$$ coordinates, which gives the coordinates of point G as:$G=\frac{r}{1-rs}\left(sb+rc, sh\right)$In the same way, the coordinates of point H are determined from the intersection of the vectors FB and AE as:$H=\frac{s}{1-rs}\left(sb+rc, sh\right)$The intersection of the vectors CD and FB gives point I’s coordinates as:$I=\frac{1}{1-rs}(r^2b+s^2c,r^2h)$
We now have the heights of the points G, H and I above the baseline AC so we can calculate the areas of triangles ACG, AFH and FCI. If we let the symbol $$\bigtriangleup$$ stand for the area of a triangle divided by the area of the red triangle ABC, we can now show that:$\bigtriangleup_{ACG}=\frac{rs}{1-rs}$ $\bigtriangleup_{AFH}=\frac{s^3}{1-rs}$$\bigtriangleup_{FCI}=\frac{r^3}{1-rs}$ We can now calculate the area of the green triangle $$\bigtriangleup_{AFH}-\bigtriangleup_{ACG}+\bigtriangleup_{FCI}$$ as (noting that $$r^2+s \equiv s^2+r\equiv 1-rs$$):$\frac{1-4rs}{1-rs}$ Setting $$r=1/3$$ and $$s=2/3$$ shows that the area of the green triangle is one seventh of that of the red triangle.