A glass suncatcher is in the shape of an equilateral triangle with sides of length $144\text{mm}$. The triangle is labeled $\triangle \rm ABC$ and divided
into $8$ smaller sections as follows.
- Sides $\rm AB$ and $\rm BC$ are each divided into $8$ segments of equal length.
- Each point of division on $\rm AB$ is connected to its corresponding point of division on $\rm BC$, creating $7$ line segments.
- Each of the $7$ line segments is parallel to the third side of the triangle, $\rm AC$.
Proposed as the problem of the week #25 of year 2025/26, by the Centre for Education in Mathematics and Computing of the University of Waterloo.
Let ${\rm X}_i$ be the $i$th point of division of the side $\rm BC$, let ${\rm Y}_i$ be that of $\rm AB$, and let ${\rm Z}_i$ be the $i$th intersection –
starting from the bottom – of $\rm AD$ with ${\rm X}_i {\rm Y}_i$; we thus want to find $[\rm Y_5 Y_6 Z_2 Z_1] : [\rm X_5 X_6 Z_2 Z_1]$.
Notice that all of the triangles of the form $\triangle {\rm D} {\rm X}_j {\rm Z}_{j-4}$ (for $j > 4$) are similar due to Thales' theorem and point number 3
in the statement, and in particular $\rm \triangle D X_6 Z_2 \sim \triangle D C A$; hence, because $\rm DX_6 = DC/2$,
$$
\rm [\triangle D X_6 Z_2] = \frac 1 4 [\triangle D C A] = \frac 1 8 DC \ DA = \frac 1 8 72 \cdot 144 \sin 60^\circ = 648\sqrt 3 \text{mm}^2.
$$
By an analogous reasoning we can compute $[\rm D X_5 Z_1]$ and find $[\rm X_5 X_6 Z_2 Z_1]$ as the difference between the two:
$$
\rm [\triangle D X_6 Z_2] = \frac 1 {16} [\triangle D C A] = 162\sqrt 3 \text{mm}^2
\implies \color{green} \boxed{[\rm X_5 X_6 Z_2 Z_1] = 486\sqrt 3 \text{mm}^2.}
$$
To obtain $[\rm Y_5 Y_6 Z_2 Z_1]$, we now look at triangles of the form $\triangle {\rm A} {\rm Y}_k {\rm Z}_{k-4}$ – which are again similar one another by the same arguments discussed above – and realize that it again can be obtained as the subtraction of two of them, namely $\triangle \rm A Y_5 Z_1$ and $\triangle \rm A Y_6 Z_2$. Therefore: $$ \begin{align*} \rm [\triangle A Y_5 Z_1] &= \frac 9 {16} [\triangle A Y_4 D] = \frac 9 {32} Y_4 Z_2\ DA = 729 \sqrt{3} \text{mm}^2,\\ \rm [\triangle A Y_6 Z_2] &= \frac 1 4 [\triangle A Y_4 D] = 324 \sqrt{3} \text{mm}^2; \end{align*} $$ and putting everything together: $$ [\rm Y_5 Y_6 Z_2 Z_1] = [\triangle A Y_5 Z_1] - [\triangle A Y_6 Z_2] = \color{green} \boxed{405 \sqrt 3 \text{mm}^2.} $$
Finally, the requested ratio turns out to be $405/486 = 5/6$.
To obtain $[\rm Y_5 Y_6 Z_2 Z_1]$, we now look at triangles of the form $\triangle {\rm A} {\rm Y}_k {\rm Z}_{k-4}$ – which are again similar one another by the same arguments discussed above – and realize that it again can be obtained as the subtraction of two of them, namely $\triangle \rm A Y_5 Z_1$ and $\triangle \rm A Y_6 Z_2$. Therefore: $$ \begin{align*} \rm [\triangle A Y_5 Z_1] &= \frac 9 {16} [\triangle A Y_4 D] = \frac 9 {32} Y_4 Z_2\ DA = 729 \sqrt{3} \text{mm}^2,\\ \rm [\triangle A Y_6 Z_2] &= \frac 1 4 [\triangle A Y_4 D] = 324 \sqrt{3} \text{mm}^2; \end{align*} $$ and putting everything together: $$ [\rm Y_5 Y_6 Z_2 Z_1] = [\triangle A Y_5 Z_1] - [\triangle A Y_6 Z_2] = \color{green} \boxed{405 \sqrt 3 \text{mm}^2.} $$
Finally, the requested ratio turns out to be $405/486 = 5/6$.