Ask for a math problem How many triangles can you draw at most with 7 straight lines that do not ove

To draw the most triangles with 7 straight lines.

It is necessary to make each of the 7 lines non-parallel to each other, and each of the 3 lines does not pass through the same point.

You can draw up to 1 triangle with 3 straight lines;

4 2 = 2, 1 + (1 + 2) = 4, - up to 4 triangles can be drawn with 4 straight lines;

5 2 = 3, 1 + (1 + 2) + (1 + 2 + 3) = 10, - up to 10 triangles can be drawn with 5 straight lines;

6 2 = 4, 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) = 20, - up to 20 triangles can be drawn with 6 straight lines;

7 2 = 5, 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + (1 + 2 + 3 + 4 + 5) = 35, - up to 35 triangles can be drawn with 7 straight lines.

For each additional line, there are n-1 more intersections, and every two intersections correspond to a new triangle. Start at n 3 and explore the discoveries.

For reference, please smile.

It should be that any 3 of the 7 straight lines can form a triangle when the number is the largest.

Then according to the permutation combination. It's C73= 35 pcs.

That is, (choose one of the 7 first.) Choose one more from the 6. Choose one more from the 5. But there are 6 repetitions. Divide by 6).

The condition is that the sail is not overlapping with any area on the hail!

Up to 11 triangles.

Three threes form 1 triangle, four lines can only form two materialism, five lines form 5, 6 lines form 7, 7 lines form 11, and so on, that is, 1 + 1 + 3 + 2 + 4 = 11.

The correct answer is 11.

Three threes form a triangle, four lines can only form two, five lines form 5, 6 lines form 7, and 7 lines form 11, which is 1 + 1 + 3 + 2 + 4 = 11.

You might as well start with 3 straight lines, and when you reach 5, you can basically understand Jingchi.

7 straight base commas draw 11 non-overlapping triangles, I drew an example in my space.

I didn't submit it just now.

The sequence should be 1 2 5 7 11 14....

The rule is that the difference is 1 (3 2) (4 3) (5 4)...My solution is quite annoying, and I haven't thought of a better way, especially when it is suitable for the exam. Because the obtained triangle has already formed the 2 3 condition of the new triangle, only a new line needs to intersect these two sides to have a new triangle generated, so the new line should be drawn as far as possible to form a triangle with the original line.

For example, the drawing method of the fifth straight line, and the number of the straight line to form a triangle of the number of the straight line to form a triangle, can add 2 triangles of the straight line skipping, generally form a non-triangle, as for the actual formation of 3 is due to the fourth straight line free intersection point, may as well be called the ** intersection point), that is, the nth straight line can be drawn after the integer number of triangles can increase (n-1) 2, the intersection of the top triangle also increases (n-1) 2 integer (** intersection point caused by the intersection of the top triangle increases by 2*** Number of intersections) 3 straight lines, 3 intersections, 1 new triangle, 1 triangle in total, 3 intersections are used, plus 0 intersections of triangles at the top, and 0 intersections remaining;

4 straight lines, 6 intersections, 1 new triangle, a total of 2 triangles, 6 intersections, plus 0+1 intersection points of the top triangle, and 1 remaining intersection points;

5 straight lines, 10 intersections, 2+1 new triangles, a total of 5 triangles, 15 intersections, plus the number of intersections of the top triangle 1+2+2=5, and the remaining intersections are 0; (Note, because there is 1 intersection left when there are 4 straight lines, and it is usually drawn before.)

Figures, straight lines are all divergent forms (intersections are already used to form triangles, such intersections directly provide a triangle, thus adding 3 triangles);

6 straight lines, 15 intersections, 2 new triangle punch heights, a total of 7 triangles, 21 intersections, plus 5+2 intersections of triangles at the top, and 1 remaining intersection;

7 straight lines, 21 intersections, 3+1 new triangles, a total of 11 triangles, 33 intersections, plus the number of intersections of triangles at the top 7+3+2=12, and the remaining intersections are 0;

8 straight lines, 28 intersection points, 3 new triangles, a total of 14 triangular scattered rulers, 42 intersection points, plus the number of intersection points of the top triangle 12 + 3 = 15, the remaining intersection point 1;

9 straight lines, 36 intersections, 4+1 new triangles, a total of 19 triangles, 57 intersections, plus 15 + 4 + 2 = 21 intersections of the top triangles, and 0 remaining intersections. Therefore, I choose 11.

The number series formed is 1 2 5 7 11 14 19....

The rule is that the difference is 1 (3 2) (4 3) (5 4)...The number of new triangles each time = [(number of lines - 1) 2] + the number of remaining intersections of the previous time.

The correct answer is 11.

Three threes form one triangle, four lines can only form two, five lines form five, six lines form seven, seven lines form 11, and so on that is, 1 + 1 + 3 + 2 + 4 = 11.

You might as well start with 3 straight lines and start with the beginning of the conversation, and draw 5 lines to basically understand.

If you choose any 3 of the 7 lines, the wheel is 35 (the combination of the most simple questions such as the wheel, I don't know if you have learned).

7 Mathematics Life 3 Page 4-7 Questions, The intersection of three different straight lines on a plane can form the logarithm of the apex angle at most.

The correct answer is 11.

Three threes form a triangle, four lines can only form two, five lines are configured into 5, 6 lines form 7, 7 lines form 11, and so on, that is, 1 + 1 + 3 + 2 + 4 = 11 You may wish to start drawing from 3 straight lines, and draw 5 is basically understood.

The condition is that there is no overlap in any area!

Up to 11 triangles.

Three sails and three groups of hail form a triangle, four lines can only form two, five materialist lines form 5, 6 lines form 7, 7 lines form 11, and so on, that is, 1 + 1 + 3 + 2 + 4 = 11.

The first three are one, and then the four are a few, and so on to find the pattern.