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# The Dollar Bill

## Question

Bill has five  pennies .  He shall make as many  triangles as he can to gain more money. For every unique triangle that he creates on a flat table (3 corners at centers of coins) he wins a nickle but for some triangles that dont overlap or intersect with other triangles, he wins a dime. He shall declare the triangles he formed on the table. So Bill can name the identical coins A,B,C,D and E. Note U\$  penny  is 1-cent, nickle is 5-cents and dime is 10-cents.

How shall Bill put all his coins on the table for greatest earnings?

## 12 answers to this question

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I appreciate TSLFs careful language.

Set 3 pennies flat on the table, and stand 2 pennies on their edges, in such a way that no four pennies are co planar. Now all  ten triangles intersect only at endpoints, so \$1.00 to Bill.

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Total of  \$ 0.70 ;  Arrange 4 pennies in a circle around a penny in the center  (A )  such that none of the  outside pennies are on the same diameter passing  thru (A)    Let's assume that as we look at our screen (B) is directly below (A) at a distance of the radius ;  Moving CW up to the left more than 90 deg is (C) ; continuing CW approx 70 degrees  which stops us short of being diametrically opposite (B) is location of (D) :  continue CW about 90 degrees for  (E).    You have 4 triangles which do NOT over lap  ABC;  ACD ; ADE ;  AEB .   You then have 6 triangles created by  lines from(B) to (D) and from (C) to (E)   These triangles being BCD ;  BDE;  BDA;  CDE ; CEA ; CEB.   So we have 4 @ 0.10 = \$040  and 6 @ .05 = 0.30 for a total of \$ 0.70

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nice but not the most that he can earn

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total of  \$ 0.75 ;  Arrange 4 pennies in a circle around a penny in the center  (A )  such that none of the  outside pennies are on the same diameter passing  thru (A)    Let's assume that as we look at our screen (B) is directly below (A) at a distance of the radius ;  Moving CW up to the left more than 90 deg is (C) ; continuing CW approx 70 degrees  which stops us short of being diametrically opposite (B) is location of (D) :  continue CW about 90 degrees for  (E).    You have 4 triangles which do NOT over lap  ABC;  ACD ; ADE ;  AEB .   You then have 6 triangles created by  lines from(B) to (D) and from (C) to (E)   These triangles being BCD ;  BDE;  BDA;  CDE ; CEA ; CEB.   So we have 4 @ 0.10 = \$040  and 6 @ .05 = 0.30 for a total of \$ 0.70 plus the 5 pennies = 75 cents

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Nice try adding the 5 cents. Note that  anyway we put the 5 pennies flat on the flat table as long as there be no 3 centers that are on the same line we can create 10 triangles. When the non intersecting triangles are declared first for dime pays then the others for nickle pays it is just luck that the dime triangles werent forfeited because the nickles triangles overlapped them. Nevertheless,  Bill can take home \$ 0.8 ( his pennies included) if he would be contented.

Edited by TimeSpaceLightForce

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Spoiler

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@ CE -Correct 3D.. so how to make the center of the 5th penny inside the tetrahedron formed by other 4 pennies?

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Place two pennies flat on the table, one of them 5 inches East of the other.

Place one penny on its edge on the West penny, on the North side.

Place one penny on its edge on the East penny, on the South side.

Place one penny on its edge, on the table between the East and West pennies.

Edited by CaptainEd
Clarify location of fifth penny

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Spoiler

coins

all coins touches the table but  WSN mess with WCE

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Ok here’s a different layout:

assume all coins have radius 1

Flat on table, centers at: A (-2, 0, 0), B: (0, 2,  0), C: (2, 0, 0)

standing on edge, centers at: D:( 0, 1, 1), E:(0, 0, 1)

no 4 coins are co-planar (that was the problem with TSLF’s picture), so all triangles meet only at endpoints. Bill gets to declare all 10 triangles.

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@CaptainEd- green triangle mess with cyan triangle

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On 9/22/2018 at 7:41 AM, CaptainEd said:

linear triangle solution

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Set 3 pennies flat on the table, and stand 2 pennies on their edges, in such a way that no four pennies are co planar. Now all  ten triangles intersect only at endpoints, so \$1.00 to Bill.

Spoiler

Your solution is simply right! . I just realized that in OP .. triangles with 3 centers connected by 3 lines can't intersect lines that way.  No 4 co planar center.

Edited by TimeSpaceLightForce

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