In the third year of junior high school, the master is advanced, and there are chasing points

Questionable title.

First, there are two scenarios.

That is, the distribution of two triangles.

Second, if it is just bac=120°AD, it is not necessarily bisected.

Add the conditions of equilateral triangles and you're fine.

But adding it seems too simple.

Wait, I'll do the math for you, have you learned trigonometric functions in the third year of junior high school, but you don't have the production sum difference theorem and formula??

I'm a junior high school student, so it's possible to kill all my brain cells. 1、 (1 2 3 … n) -n-1)*250/7 = n(1 n)/2 - n-1)*250/7 (250/7) *2=

26 First Question.

First of all, the OAB triangle is equilateral, so the length of the three sides of OA, OB, and AB is 2.

The question is to ask about the area of the triangle opq in relation to time t.

Point Q is 1 unit per second moving from O to C, while P is 3 units per second from A to O and then to B.

Let's take a look at point P, when it moves to point O, the area OPQ is only related to the triangle of AOC, but after point O, it is more complicated, so let's consider the relationship between point P and point O:

When does point P move from point A to point O?

Remember that ABO is an equilateral triangle, so the length of AO is 2, so the distance from point A to point O is also 2.

The topic also said that p moves 3 units per second, so at 2 3 seconds, point p has already moved from point A to point O.

So the first case is 0 t 2 3 (from point A to point O).

Now let's look at point Q, how much time does it take for point Q to move from point O to point C?

To know this answer, we first need to calculate the distance of the OC.

OCA is known to be an isosceles triangle, i.e., OC AC, and angle C is equal to 120.

According to the theory of isosceles triangles, the angle o is equal to the angle a, and according to the theory of triangle sum 180, the angle o and the angle a are both 30 degrees.

And then we cut the triangle in half at angle c, so we have two triangles of 30 degrees, 60 degrees, 90 degrees. (120 divided by 2 60).

Since OA 2, the cut half is 1, which is the side of 30 degrees and 90 degrees.

According to the triangle of 30 degrees, 60 degrees and 90 degrees, the side of oc is 2 square 3, and after simplification, 2 square 3 3

In other words, it takes 2 to 3 3 seconds for Q to move from O to C.

So the second case is 2 3 (after case 1) t 2 open 3 3

In case 3, it is natural that Q goes from O to C, and P reaches point B.

The distance from A to O to B at point P is 4, and it moves 3 units per second, so the total time required is 4 3 seconds, so case 3 is 2 square 3 3 t 4 3 (you didn't write the answer, and I don't know if you need to count this).

26。(1) oc=2 3 3, so the motion time of q is 2 3 3, and the motion time of p is 4 3>2 3 3, because q stops, p also stops, so t<=2 3 3When p passes the o point, the height and bottom of the triangle change, and at this time t = 2 3.

2) D is on the edge of a triangle, when D is on OA, there are two cases, one is OD=OC=2 3 3, so D(2 3 3,0), the other is OD=DC, the height on the OC side, the right triangle with a 30 degree angle, OD=2 3, D(2 3,0), when D is on OB, only OC=OD, through D as the perpendicular line of the X axis, and also use the right triangle with a 30 degree angle, D(3 3,1), when D is on AB, it is a bit troublesome, as the perpendicular line of OC, The intersection with ab is d, which I don't really want to ask for, and I forget many formulas. See for yourself.