MathStatClub
PROBLEM OF THE MONTH
March 2015
Problem: Consider a unit circle called C1 and another circle of radius r, called Cr , where
r 6= 1. For which values of r can these two circles be arranged so that the arc of C1 enclosed
by Cr is exactly twice as long as the arc of Cr enclosed by C1 ? Give a convincing argument
for your claim.
Please submit your solution to:
• Dr. Marko Samara, [email protected]
before the deadline: March 31, 7:00PM. The WINNER will be awarded with a $15 dining
gift card and certificate, and will be announced in the next issue of the Problem of the
Month.
Winner of the February’s problem: Allison Wedegis
Other students with correct answer:
Samuel Dillow
Luis Matos
Due Nguyen
Matthew McCusker
Qifei Hao
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(Congratulations!)
Solution to the February’s Problem of the Month
Problem: Suppose you have twelve golden coins, all the same in shape and size. However,
one of them is fake: it does seem golden from the outside, but its core is made of a cheaper
material, so its weight differs from the weight of others (all the others weigh equally). Using
the scales as in the picture below, and taking at most three measurements, how can you
find which coin is fake AND whether it is lighter or heavier than the others? You need to
describe a procedure/algorithm so that whatever happens in each of the three measurements,
the procedure will in any case lead to the answer.
Solution:
There is more than one algorithm, all similar, and with the same first step. We show one of
them.
In the first measurement, take four coins on one scale, other four coins on the other scale,
and set the remaining four coins aside.
Case 1: The scales (in the first measurement) are in balance.
In this case, one of the remaining four coins set aside is fake. Denote these coins by A, B,
C and D. The rest of the coins are ”good”. For the second measurement, place A and B on
one scale, and C with one of the good coins on the other scale.
Case 1a: The scales (in the second measurement of case 1) are in balance.
In this case, the coin D is fake, and all other 11 are good. In the third measurement, place
coin D on one of the scales, and any one of the good coins on the other scale. If D goes down,
D is fake AND is heavier than all others. If D goes up, D is fake AND is lighter than all others.
Case 1b: The scale with A and B (in the second measurement of case 1) dips down, while
the scale with C and the good coin goes up.
In this case, either A or B is fake and heavier, OR C is fake and lighter. The rest of the coins
are all good. In the third measurement, place the coin A on one scale and the coin B on the
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other. If the scale is in balance, we conclude the coin C is the fake one AND is lighter than
all the others. If the scale with coin A dips down, we conclude A is the fake one AND is
heavier (note that such a result of the third measurement cannot be due to B being fake, as
if it was fake, it could have been heavier, and could not go up). If, however, the scale with
the coin A goes up (i.e. the scale with B goes down), we conclude the coin B is the fake one
AND is heavier than the others.
Case 2: One of the scales (in the first measurement) goes down, the other one goes up.
In this case we have four coins in the down scale being potentially fake and heavier (denote
them by H1 , H2 , H3 , H4 ), and four coins in the up scale being potentially fake and lighter
(denote them by L1 , L2 , L3 , L4 ). The remaining four coins (set aside) are all good. In the
third measurement, place H1 , H2 , L1 on one scale, and H3 , H4 , L2 on the other scale
Case 2a: The scales (in the second measurement of case 2) are in balance.
This means one of the coins L3 , L4 is fake and lighter. So, in the third measurement, place L3
on one scale and L4 on the other. Apparently, the scales will not be in balance. Whichever
of the coins L3 , L4 goes up in the third measurement - is the fake coin AND is lighter.
Case 2b: The scale with coins H1 , H2 , L1 (in the second measurement of case 2) goes
down.
This means either one of the coins H1 , H2 is fake and heavier OR, coin L2 is fake and lighter.
So, in the third measurement, put H1 on one scale and H2 on the other. If the scales are
balanced, we conclude the fake coin is L2 , AND it is lighter than the others. If, however,
the scales are unbalanced, whichever of H1 , H2 went down - is the fake coin, AND is heavier
than the others.
Case 2c: The scale with coins H3 , H4 , L2 (in the second measurement of case 2) goes down.
This case is symmetric to the case 2b: just repeat the procedure as in case 2b, with H3 , H4
replacing H1 , H2 , and L2 replacing L1 .
The described procedure/algorithm covers all possible cases.
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