The cat can't swim, so the mouse is safe in the water for the time being, but the mouse will drown if it doesn't go ashore, so it still has to find a way to get ashore before the cat catches up with it. When it goes ashore, there will be a mouse hole. Haha, if the mouse swims directly along the radius, it will definitely die on the cat's paw. T 1=R/V,

Cat time T2 = ∏ r/4v = (3.14) r/v, t 1 > T2, the cat is waiting for the mouse on the shore. But this mouse is unusual. It took part in the China Mathematics Competition. The mouse first watched the cat move. If the cat runs clockwise (counterclockwise) along the shore, the mouse swims clockwise around the middle of the lake. As long as the mouse swims within the radius of R/4 around the center of the lake, it can keep a confrontation with the cat, that is, the cat, the center of the lake and the mouse keep a straight line, because the distance that the mouse swims at this time is smaller than that of the cat. ，(∏R/4:∏R=？ ), that is to say, rats have an advantage in a small circle with a radius of R/4, at least even. At a certain moment, the mouse just swam to a small circle with a radius of R/4, crossed the middle of the lake and faced the cat. Knowing that the opportunity came, the mouse immediately turned around and swam to the shore along the diameter, using T 1=(3/4)R/V, so the cat was blindsided, and it used T2 = ∏ R/4V = (3. 14. T2, the mouse went home.