The following problem was posed in the March 2025 issue of Crux Mathematicorum.
Solution to MA311 by Srikanth and Achudhan.
Show Solution
Let $l$ be the sidelength of the equilateral triangle $\triangle DFG$. Since $\angle E F G=\angle E B G$, the points $E F B G$ are cyclic, we have $\angle GEF =90^{\circ}$. Thus $GE$ is an altitude of the equilateral triangle and hence $E$ is mid-point of $FD$. So $$FE = l/2.$$
Next, note that $\triangle ADF \cong \triangle CDG$ since the sides of the square (as hypotenuse), the sides of a common equilateral triangle and a right angle are present in both triangles. So we conclude $\angle CDG = 15^{\circ}$. Since $\angle EDG = 60^{\circ} = \angle ECG$, the points $E,D,C,G$ are concyclic, and thus $\angle GED = 15^{\circ}$. This forces $\angle GEB = 45^{\circ}$ and thus $\angle GFB = 45^{\circ}$, since $EFBG$ is also concyclic. By angle sum property in $\triangle FBG$, we conclude its an isosceles triangle with $$BG = \frac{FG}{\sqrt{2}}=\frac{l}{\sqrt{2}}.$$ Since angle at H is vertically opposite and $\angle EFH = \angle HBG$, we have $\triangle EFH$ is similar to $\triangle GBH$. Thus we have $$\dfrac{[EFH]}{[GBH]} = \dfrac{EF^2}{BG^2} = \dfrac{l^2/4}{l^2/2}=\dfrac12.$$
Authors:
R.Achudhan is a 9th grader who is enrolled in Target RMO 25 at Mudhitha Maths Academy ,Chennai. He loves solving problems especially non-routine kind. His love for mathematics and problem solving has kindled his journey into Mathematics Olympiads under the guidance of the Academy. He has also cleared IOQM 24-25 and secured 5th rank in NMTC contest.
Srikanth is the instructor at Mudhitha Maths Academy, Chennai. He is very fond of problem solving and puzzle games. He is an assistant professor of mathematics by the week, enjoys designing software and loves teaching mathematics. Srikanth likes to play role playing or strategy video games, enjoys reading fantasy novels and collect fountain pens. His favorite subjects are geometry and category theory.