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Solution to the final problem of mathematics in senior high school entrance examination
Question (1): The center of the circle is at the midpoint of DE, and the arc is tangent to BC. The result is the length of a semicircle with a diameter of DE.

The first short question in question (2) takes the result of question (1) as the boundary reference:

First of all, the center of circle P can only be on the vertical line of Germany, and the inner arc formed by Germany can only be downward as shown in the question.

Secondly, to ensure the formation of the middle inner arc, the distance from point P to BC line must be ≥ the distance from point P to point D (or E).

Combining the results of the problem (1), we can know that when t= 1/2, the ordinate of point p is between 1 and 1.5 (inclusive).

The second question is the concrete calculation based on the first question.

In principle, the distance from P point to BC line must be ≥ the distance from P point to D (or E).

So where is the limit position of point P?

According to the result of the first question, we can know that if point P is the limit position at (t, 1.5), it is most likely that there is an intermediate inner arc, and if it does not exist at this time, it is even more impossible to have other positions.

√(t^2+0.5*0.5)≤ 1.5

Get 0 < t ≤√ 2.