## - Problem 3

Let $A B C D$ be a convex quadrilateral. Let $P$ be the intersection of $A C$ and $B D$. The angle $\angle A P D=60^{\circ}$. Let $E, F, G$ and $H$ be the midpoints of the sides $A B, B C, C D$ and $D A$ respectively. Find the largest positive real number $k$ such that

$$
E G+3 H F \geq k d+(1-k) s
$$

where $s$ is the semiperimeter of the quadrilateral $A B C D$ and $d$ is the sum of the lengths of its diagonals. When is equality achieved?
## Solution:

We will prove that $k=1+\sqrt{3}$ and that equality holds if and only if $A B C D$ is a rectangle.

![](https://cdn.mathpix.com/cropped/2024_04_26_70e03d44ef1ac75a53e9g-2.jpg?height=614&width=1005&top_left_y=1486&top_left_x=571)

Let $W, X, Y$ and $Z$ be four points outside $A B C D$ such that triangles $A B W$ and $D C Y$ are equilateral, triangle $B C X$ is isosceles at $X$, triangle $A Z D$ is isosceles at $Z$, and $\angle B X C = \angle A Z D = 120^{\circ}$.

The quadrilaterals $W A P B, X B P C, Y C P D$ and $Z D P A$ are cyclic. Therefore, by Ptolemy's theorem, we obtain:

$$
W P = P A + P B, \quad X P \sqrt{3} = P B + P C, \quad Y P = P C + P D, \quad Z P \sqrt{3} = P D + P A
$$

On the other hand,

$$
\angle W P Y = \angle W P B + 60^{\circ} + \angle C P Y = \angle W A B + 60^{\circ} + \angle C D Y = 180^{\circ}
$$

Thus, $W, P, Y$ are collinear and, similarly, $Z, P, X$ are collinear. Therefore:

$$
\begin{aligned}
W Y & = W P + P Y = P A + P B + P C + P D = A C + B D \\
X Z & = X P + P Z = \frac{1}{\sqrt{3}}(P B + P C + P D + P A) = \frac{1}{\sqrt{3}}(A C + B D)
\end{aligned}
$$

By the triangle inequality:

$$
W Y \leq W E + E G + G Y, \quad X Z \leq X F + F H + H Z
$$

Thus:

$$
A C + B D \leq A B \frac{\sqrt{3}}{2} + E G + D C \frac{\sqrt{3}}{2}, \quad \frac{1}{\sqrt{3}}(A C + B D) \leq \frac{B C}{2 \sqrt{3}} + F H + \frac{A D}{2 \sqrt{3}}
$$

Therefore, summing up,

$$
\begin{aligned}
A C + B D & \leq A B \frac{\sqrt{3}}{2} + E G + D C \frac{\sqrt{3}}{2} \\
\sqrt{3}(A C + B D) & \leq B C \frac{\sqrt{3}}{2} + 3 F H + A D \frac{\sqrt{3}}{2} \\
\hline(1+\sqrt{3})(A C + B D) & \leq E G + 3 F H + s \sqrt{3}
\end{aligned}
$$

That is,

$$
E G + 3 F H \geq (1+\sqrt{3}) d - s \sqrt{3}
$$

Thus, if $k=1+\sqrt{3}$, then $E G + 3 F H \geq k d + (1-k) s$.

Equality holds if and only if, on one hand, $W, E, G, Y$ are collinear and, on the other hand, $X, F, H, Z$ are also collinear. Since $W E$ is perpendicular to $A B$ and $G Y$ is perpendicular to $D C$, $A B$ and $D C$ must be parallel, and similarly, $B C$ and $A D$ must also be parallel, so $A B C D$ must be a parallelogram. Additionally, the line $E G$ is perpendicular to $D C$, which implies that $A B C D$ is a rectangle, and it is easily verified that if $A B C D$ is a rectangle, then equality holds. Therefore, equality holds if and only if $A B C D$ is a rectangle.

Now, let $l$ be a positive real number such that $E G + 3 H F \geq l d + (1-l) s$. Then, if $A B C D$ is a rectangle,

$$
k d + (1-k) s \geq l d + (1-l) s
$$

That is,

$$
k(d-s) \geq l(d-s)
$$

But the triangle inequality implies that $d > s$, which implies that $k \geq l$. Therefore, the real number sought is $k=1+\sqrt{3}$ and equality holds if and only if $A B C D$ is a rectangle.