Problem 1695

1. Setup and Definitions:
Let $ABCD$ be a trapezoid with bases $AB$ and $CD$ such that $CD = k \cdot AB$ where $0 < k < 1$. Point $P$ is such that $\angle PAB = \angle CAD$ and $\angle PBA = \angle DBC$. We need to prove that $PA + PB \leq \dfrac{1}{\sqrt{1-k^2}} \cdot AB$.

2. Introduce Perpendiculars:
Let $X, Y \in DB$ be such that $AX$ and $BY$ are perpendicular to $DB$. Let $XA = YB = h$, $XD = a$, $YB = b$, $AB = c$, and $CD = kc$. Moreover, let $\angle PAB = \angle DAC = x$ and $\angle PBE = \angle DBC = y$.

3. Trigonometric Inequality:
We need to prove that $PA + PB \leq \dfrac{1}{\sqrt{1-k^2}} \cdot AB$, which translates to $\sin x + \sin y \leq \dfrac{1}{\sqrt{1-k^2}} \sin(x+y)$. Since $\sin(x+y) = \sin x \cos y + \sin y \cos x$, we need to show that:
$$\sqrt{1-k^2}(\sin x + \sin y) \leq \sin x \cos y + \sin y \cos x$$

4. Claim 1:
$$\sin x = \dfrac{DC \cdot h}{AD \cdot AC} \quad \text{and} \quad \sin y = \dfrac{DC \cdot h}{BD \cdot BC}$$
Proof: We will only prove the first equality, as the proof for the second one is similar. Note that:
$$\dfrac{AD \cdot AC \sin x}{2} = (ADC) = \dfrac{DC \cdot h}{2}$$
Hence, we obtain the desired equality:
$$\sin x = \dfrac{DC \cdot h}{AD \cdot AC}$$
$\blacksquare$

5. Rewriting the Inequality:
By our Claim, we are equivalently left to prove that:
$$\sqrt{1-k^2} \cdot \left( \dfrac{1}{AD \cdot AC} + \dfrac{1}{BD \cdot BC} \right) \leq \dfrac{\cos y}{AD \cdot AC} + \dfrac{\cos x}{BD \cdot BC}$$
That is:
$$\sqrt{1-k^2}(BD \cdot BC + AD \cdot AC) \leq BD \cdot BC \cos y + AD \cdot AC \cos x$$

6. Claim 2:
$$BD \cdot BC \cos y + AD \cdot AC \cos x = AD^2 + BC^2 + AB \cdot CD - CD^2$$
Proof: Let $P \in AD$ and $Q \in BC$ be such that $CP$ is perpendicular to $AD$ and $DQ$ is perpendicular to $BC$. Note that:
$$BD \cdot BC \cos y + AD \cdot AC \cos x = BQ \cdot BC + AD \cdot AP = BC^2 + AD^2 + QC \cdot BC + PD \cdot DA$$
$$= BC^2 + AD^2 + DC \cdot CY + DC \cdot CX = AD^2 + BC^2 + (DY + CX) \cdot DC = AD^2 + BC^2 + AB \cdot CD - CD^2$$
$\blacksquare$

7. Final Inequality:
Therefore, returning to the problem, we need to prove that:
$$AD^2 + BC^2 + AB \cdot CD - CD^2 \geq (AD \cdot AC + BD \cdot BC) \sqrt{1-k^2}$$
That is (switching to variables and heavy algebra):
$$2h^2 + a^2 + b^2 + (k - k^2)c^2 \geq \sqrt{1-k^2} \left( \sqrt{(h^2 + a^2)(h^2 + (c-b)^2)} + \sqrt{(h^2 + b^2)(h^2 + (c-a)^2)} \right)$$
Note that we have the condition $a + b = c - kc$, as $a + b = XA + CY = XY - DC = AB - DC = c - kc$.

8. Cauchy-Schwarz Inequality:
By the Cauchy-Schwarz inequality, it suffices to prove that:
$$(2h^2 + a^2 + b^2 + (k - k^2)c^2)^2 \geq (1-k^2)((h^2 + a^2) + (h^2 + b^2))((h^2 + (c-b)^2) + (h^2 + (c-a)^2))$$
Or equivalently (using $a + b = c - kc$) that:
$$(2h^2 + a^2 + b^2 + (k - k^2)c^2)^2 \geq (1-k^2)(2h^2 + a^2 + b^2)(2h^2 + 2kc^2 + a^2 + b^2)$$
Put $2h^2 + a^2 + b^2 = M$. Then, we are left to prove that:
$$M(M + 2kc^2)(1-k^2) \leq (M + (k - k^2)c^2)^2$$
Which is equivalent to:
$$(kM - (k - k^2)c^2)^2 \geq 0$$
This holds, hence we may conclude.

The final answer is $\dfrac{1}{\sqrt{1-k^2}} \cdot AB$