Bill was asked to form as many triangles on top of a flat table using his 5 pennies (3 coins on the table where centers are connected by imaginary lines to make a triangle). His teacher told him that for every triangle formed, Bill will be paid a nickle (5 cents) if he declares it. But, for every triangle that does not intersect with other triangle he declared, he will be given a dime (10 cents).

How should Bill place all his coins on the table for greatest earning?

  • 1
    $\begingroup$ Does he get to place the nickel and dime on the table to increase the triangle count? $\endgroup$ – Ian MacDonald Jun 20 '18 at 14:02
  • $\begingroup$ no. he will be paid after he place his coins and accounted the figures $\endgroup$ – TSLF Jun 20 '18 at 14:14
  • $\begingroup$ Does he have to declare the triangles he gets dimes for? And does he get both a nickle and a dime for a triangle satisfying both the nickle condition and the dime condition, or does he just get a dime? $\endgroup$ – user2357112 Jun 21 '18 at 18:36
  • $\begingroup$ he should not delare a triangle that cause intersection..so that he earns double for clean triangle.. possible 5 if all coins are flat on the table $\endgroup$ – TSLF Jun 21 '18 at 21:06

One Dollar

One dollar

First, make a roughly equilateral triangle by placing pennies A, B, and C flat on the table. This forms one triangle. Next, balance penny D on its edge in the center of triangle ABC. This forms three more non-intersecting triangles in the air, as the center of D is elevated relative to the centers of A, B, and C. Lastly, place penny E flat on the table, in line with pennies A and D. This forms three non-intersecting triangles in the air, each involving E and D. At this point, the four pennies flat on the table form four triangles that overlap. We can recover BCE and keep the three triangles ABC, ABE, and ACE if we can slightly lift the center of A without going over the line between the centers of A and E. To do so, simply pick up A and rest it partly on E, raising its center.

Which should look something like the following image:

enter image description here

  • $\begingroup$ 1vote for the standing coin..your british penny solution is acceptable if it can not be done with all american pennies $\endgroup$ – TSLF Jun 21 '18 at 15:50
  • $\begingroup$ @TSLF If you'll accept all British pennies, then (according to wiki), the steel penny is slightly thicker than the copper. You could replace penny A with a steel one, rather than balancing it and relying on placing E far enough out. I think this is the best you can do while still having all pennies touch the table. $\endgroup$ – Joel Harmon Jun 22 '18 at 3:02
  • $\begingroup$ I read the question's strange sounding definition again, and I think your solution works with all coins the same size too: with 2 (or 3) lying flat (and some little care taken not to place any four of the centres on the same plane), some triangles will intersect, but the intersection will be a line segment, which probably doesn't count as "intersecting area". $\endgroup$ – Bass Jun 22 '18 at 6:16
  • $\begingroup$ @TSLF That should do it for a universal solution. Let me know if I can make it more clear. $\endgroup$ – Joel Harmon Jun 23 '18 at 13:06

I think that

it's impossible to put more than three coins on the table without forming intersecting/partially overlapping triangles.


Bill should just concentrate on making as much triangles, so if he puts the five coins randomly on the table and makes sure there aren't three collinear ones, every three coins will form a triangle, so he'll get 5 over 3 = 10 triangles and earns 50 cent.

Of course,

with a bit of , he can put one coin on the table, add three coins in a layer on top of the first coin, and finish with the last coin. Looking at the configuration in three dimensions, none of the 10 triangles are intersecting and he'll get 10 * 10 = 100 cents = a Dollar Bill.

  • $\begingroup$ Bill can declare fewer triangles $\endgroup$ – TSLF Jun 20 '18 at 14:20
  • 2
    $\begingroup$ That's not what the question states right now, it says "formed". $\endgroup$ – Glorfindel Jun 20 '18 at 14:21
  • $\begingroup$ Right about 3D..but all coins should be on the table or touching the table. Note the triangle formed with first & last coin intersects with mid horizontal triangle? $\endgroup$ – TSLF Jun 20 '18 at 15:24
  • 1
    $\begingroup$ In 3D, he should just not declare the middle triangle. Then there will be no intersection of any of the 9 triangles, for a total of 90 cents. $\endgroup$ – user3294068 Jun 20 '18 at 17:57
  • $\begingroup$ If he just shifted the first coin to the top of the last coin that will be 10 triangles w/o intersect $\endgroup$ – TSLF Jun 21 '18 at 20:55

I would think the answer is:

10 triangles, or $1.00 profit declaring no triangles.

Piece 1 of logic:

Any 3 coins on the table that are not co-linear will form a triangle. To maximize his returns, Bill has to put all 5 coins on the table such that no 3 form a straight line which is quite simple.

Piece 2 of logic (lateral thinking):

Bill should not declare any triangles. The problem does not state he won't get paid 10 cents for a triangle he doesn't declare, just that he won't get paid 10 cents if it intersects one he did declare. By declaring no triangles, Bill guarantees each triangle will earn him 10 cents.

  • $\begingroup$ His geometry teacher would say Bill created a pentagon $\endgroup$ – TSLF Jun 21 '18 at 15:42

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