Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Find the answer

2. As shown in Figure 2, person A at point $A$ on the shore notices person B in distress at point $B$ in the water. Point $B$ is 30 meters away from the shore at point $C$ , and $\angle BAC=15^{\circ}$ . Person A's running speed on the shore is $\sqrt{2}$ times their swimming speed in the water. Given that person A's swimming speed in the water is 3 meters/second, the time it takes for person A to reach point $B$ from point $A$ is $t$ seconds. Then the minimum value of $t$ is $\qquad$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

2. $15 \sqrt{2}+5 \sqrt{6}$

As shown in Figure 3, construct
$\angle C A D=45^{\circ}$ , and let point $P$ be where person 甲 enters the water from the shore. Draw $P H \perp A D$ at point $H$ , at this time,
$A P=\sqrt{2} P H .$

According to the problem, the time it takes for 甲 to travel from point $A$ to $B$ is the time spent swimming the distance $P H + P B$ in the water.

Draw $B E \perp A D$ at point $E$ . By the shortest distance from a point to a line, we know $P H + P B \geqslant B E$ , with equality holding if and only if point $E$ coincides with $H$ .
In the right triangle $\triangle A B C$ ,
$A B=\frac{B C}{\sin \angle C A B}=\frac{30}{\sin 15^{\circ}} \text {. }$

In the right triangle $\triangle A E B$ ,
$\begin{array}{l} B E=A B \sin (\angle D A C+\angle C A B) \\ =45 \sqrt{2}+15 \sqrt{6} . \end{array}$

Therefore, $t_{\min }=15 \sqrt{2}+5 \sqrt{6}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.