Is it possible that the clock face of the clock is divided into three equal areas by three needles?

Impossible (if this problem is a brain teaser, then the answer is that the clock is broken) Solution: Timekeeping starts at 00:00 Let x seconds be divided into thirds x<86400 The hour hand travels 1° 120 per second, the minute hand moves 1° per second, the 10 second hand moves 6° per second, the hour hand and the second hand differ by 120°, then 6x-x 120-m360= 120, where m is an integer x=( 120+m360) 120 719 The hour hand and the minute hand are 120° apart, then x 10-x 120-n360= 120 where n is an integer x=( 120+n360) 120 11 Since the denominator of x obtained by the two formulas (719 and 11) is coprime The intersection of the two formulas cannot be found in x<86400.

Calculate n times in an hour, and 24n is what you want. So how to calculate an hour Take it slowly: first count the hour hand and the minute hand, take the hour hand as a reference, the minute hand is at an angle of 120 degrees to him, there are two chances, according to the current 12 o'clock, then, the first time is in the past:

120 (360-30) = formed after 12/33 (hour) ("360" and "30" are respectively the degree of the hour and minute hand revolutions, i.e., the speed) At this time, the position of the second hand is 360 times 60 times 12/33 and then divided by 360, and the remainder position is about 294 degrees, and the hour hand is not turned by one degree, then the three hands are not on the minute side. According to this idea, the chance of not having three stitches in 24 hours can be solved!! Therefore, the topic is unsolved, and I thank you.

3 o'clock is 90 degrees.

The minute hand travels 6 degrees per minute, and the hour hand travels every minute.

Start at 3 o'clock.

Let the hour and minute hands be at an equal distance from the "3" after the passage of x minutes, and be on either side of the "3".

x=180/13

That is, at 3:13 and 11/13, the hour and minute hands are at the same distance from the "3", and are on either side of the "3".

= 13 and 11 13 points.

Let the hour and minute hands be at the same distance from the "3" and be on either side of the "3" when 3 is past x minutes on the clock face.

30×3-6x=

x=180/13

x = 13 and 11 13

When 3 o'clock passes 13 and 11 13 minutes on the clock face, the hour and minute hands are at equal distances from the "3" and are on either side of the "3".

This kind of question is super simple, you can't do it, I really admire, boy! Hee-hee-hee-hee...

At 3 o'clock, the hour hand points to 3 and the minute hand points to 12The minute hand is 15 stops behind the hour hand. To coincide the minute hand with the hour hand, which translates into the minute hand chasing the time hand, the speed of the minute hand is 1 block per minute, and the speed of the hour hand is 1/12 block per minute, so 15 (1 1 12) = 15 12 11 = 15 (1 + 1 11) = 16 + 4 11 (minutes), and the minute hand coincides with the hour hand at 3 hours (16 + 4 11) minutes.

Draw two parallel lines on the circle and divide the circle into three parts: upper, middle, and lower: upper: 11 12 1 2;Medium:

9 10 3 4；Bottom: 8 7 6 5;And both are 26. Add up the numbers 1 to 12 on the clock, and the sum is 78, because the values of the three parts should be equal, so divide 78 by 3 to equal 26, as shown in the following figure:

When 3 points crack and slow dust, the minute hand and the hour hand are 15 squares apart, the minute hand goes one square, and the hour hand goes 1 12 squares 15 (1-1 12) = 16 and 4 11 minutes are perpendicular to each other, plus 15 squares 30 (1-1 12) = 32 and 8 11 minutes Answer: On the clock face 3:16 and 4 11 minutes, the minute hand and the hour hand coincide 3:32 and 8 11 minutes on the clock face ,..

Set 3 hours x minutes on the clock face, and the minute hand coincides with the hour hand.

90 + solution: x = 180 11

Answer: At 3:180 11 on the clock face, the minute hand coincides with the hour hand.

At 3 o'clock, the minute hand and the hour hand are 15 squares apart, the minute hand goes one square, the hour hand goes 1 and the hour hand goes 12 squares.

15 (1-1 12) = 16 and 4 11 points.

Perpendicular to each other, add another 15 blocks.

30 (1-1 let pure 12) = 32 and 8 11 points.

Answer: At 3:16 and 4:11 on the clock face, the minute hand coincides with the hour hand.

At 3:32 and 8 11 on the clock face, the minute and hour hands are exactly perpendicular to each other.

Let 3 o'clock exceed x minutes on the clock face, and the hour and minute hands are at equal distances from the "3" and on either side of the "3".

15-x=x/12

x/12+x=15

13/12x=15

x=180/13

So on the clock face 3 o'clock more than 180 13 minutes, the hour and minute hands are at equal distances from the "3" and are on either side of the "3".

Solution: According to the laws of clocks and watches, it is easy to know

The minute hand travels 6° per minute, and the hour hand passes every minute.

Let the hour and minute hands be at the same distance from 3 at 3 on the clock face, and on either side of 3, there is:

90°-6°x=

Solution: x minutes, 13 minutes, 50 seconds, 46 seconds

Answer: At 3:13:50.46 on the clock face, the hour and minute hands are at equal distances from 3 and are on either side of 3. Finished.

Answer. Let the clock face exceed 3 o'clock and x minutes, and the hour and minute hands are at equal distances from "3", and on both sides of "3", according to the title.

15-x=1/12ⅹ

15-x+x=1/12ⅹ+x

15x12 13=12 13xx12 13 This is 3:13 and 11 13.

Answer: On the clock face, after 3:13 and 11 13 minutes, the hour and minute hands are at equal distances from the "3" and are on either side of the "3".

Parse. When the hour and minute hands on the clock face are on either side of "3" and the distance from "3" is equal, the minute hand should be in front of 3, the hour hand should be after 3, when 3 o'clock is the hour hand pointing to 3, the minute hand is pointing to 12, and the grid between the minute hand and the hour hand is 15 squares, according to: 15-the number of squares traveled by the minute hand = 1 12 The time of the minute hand can be solved.

Comments. The key to this problem is to solve the equation according to the equal distance of the hour and minute hands from "3".

Every 60 minutes.

The hour hand goes 30 degrees.

The corresponding minute hand travels 360 degrees, and when 3 o'clock x minute, the distance between the hour and minute hands and 3 is equal.

And on both sides of the 3.

At this time. The angle of the hour hand distance at 3 o'clock is.

30x/60=x/2

The angle at which the minute hand travels is:

360x/60=6x

The angle at the minute hand distance of 3 is .

90-6x rule.

x/2=90-6x

x=180/13

At 3:180 13 o'clock, the hour and minute hands are at equal distances from 3.

And on both sides of the 3.

Let 3 o'clock exceed x minutes on the clock face, and the hour and minute hands are at equal distances from the "3" and on either side of the "3".

15-x=x/12

x/12+x=15

13/12x=15

x=180/13

So on the clock face 3 o'clock more than 180 13 minutes, the hour and minute hands are at equal distances from the "3" and are on either side of the "3".

Solution: Let the needle and minute hands be at equal distances from 3 at 3 hours n minutes, and at"3"on both sides.

then there is 15-n = n 12

The solution is n=180 13