From Newsgroup: rec.puzzles


There's a broad category of puzzles called *Find the Counterfeit coin.
You're given a large number of genuine coins and a bag of N labeled coins exactly one of which is counterfeit. The counterfeit might be heavy or light, but will expose itself if compared with a genuine coin on a balance scale.
Your objective is to establish how many weighings are needed to guarantee identifying the counterfeit.

For example, 3 weighings are enough to find the counterfeit from among
14 suspects. (If you are required to determine whether the counterfeit is heavy or light, only 13 suspects can be resolved.) In this 14-coin solution
you must vary your 2nd and 3rd weighings depending on the earlier results.
If instead your exact scale match-ups are all predetermined, you have
an "oblivious" solution.

(1) Is there an oblivious solution using only 4 weighings for 13 suspect coins?

Now let's suppose your balance scales are unreliable. During any test
sequence they may hallucinate an incorrect answer once or twice (or not at all).

(2) With two suspect coins (exactly one of which is counterfeit) there is an oblivious procedure to locate the counterfeit with five weighings (up to
two of which may report an erroneous result). Can you find such a procedure?

(The following puzzles may not be easy. I don't require that any of
these procedures be oblivious.)

(3) How many weighings (with two fictions allowed) are needed to distinguish among 3 suspect coins?

(4) How many weighings (with two fictions allowed) are needed to distinguish among 4 suspect coins?

Good Luck!
James
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From Newsgroup: rec.puzzles


Traffic in r.p was about 20 messages per week recently, but in the last two weeks the only posts are by me. :-(

Do I have B.O. or something? (I don't think you can smell me from where you are. Could it be that today's smart-phones have secret odor sensors and are sending my stats to some place on the Dark Dark Web?)

Anyway, here are spoilers for some of the weighings solutions.
I'll insert an unrelated alt.math.rec problem as spoiler space in case anyone still wants to work on the weighings.

============ Here is unrelated puzzle as spoiler space ================

"My name is James and I'm a YouTubeaholic."
I'm not overly ashamed of this addiction. It's much less expensive than my other major addiction; and, I think, preferable to being a Facebook addict
or Xwitter critter. With Firefox frequently going into irrecoverable stall
(as I experiment to find the operational maximum on the number of open tabs), I'm often presented with YouTubes I WANT to click, but avoid clicking. We know from the Manhattan project k_eff > 1 will lead to unlimited tab explosion.

I was NOT tempted to click a recent YouTube where the title gave
BOTH a problem in geometry AND its solution.

A circle inscribed in 3-4-5 triangle has area of pi.

Whippee! I set my aging brain to work and, suspecting I had overlooked
some clever shortcut, found three equations (two linear, one quadratic)
in three unknowns. Given the answer (pi) we have a 4th equation, but
they were cumbersome enough that I hardly wanted to continue. I did though; everything canceled and r = 1 = 1 was the happy result!

Feel free to post your more clever solution and show me up.

========================= End of Spoiler warning =======================



James Dow Allen <user4353@newsgrouper.org.invalid> posted:

There's a broad category of puzzles called *Find the Counterfeit coin.
You're given a large number of genuine coins and a bag of N labeled coins exactly one of which is counterfeit. The counterfeit might be heavy or light,
but will expose itself if compared with a genuine coin on a balance scale. Your objective is to establish how many weighings are needed to guarantee identifying the counterfeit.

For example, 3 weighings are enough to find the counterfeit from among
14 suspects. (If you are required to determine whether the counterfeit is heavy or light, only 13 suspects can be resolved.) In this 14-coin solution you must vary your 2nd and 3rd weighings depending on the earlier results.
If instead your exact scale match-ups are all predetermined, you have
an "oblivious" solution.

(1) Is there an oblivious solution using only 4 weighings for 13 suspect coins?

Label the 13 coins
u,
aw, ax, ay, az,
bw, bx, by, bz,
cw, cx, cy, cz,

Weigh the 4 a's against the 4 b's, and then the 4 a's against the 4 c's.
If both weighings are unequal, the counterfeit is an a.
The counterfeit is b, c, or u respectively if the first, second or neither weighing is unequal.
Now weigh the w's against the x's and, finally, the w's against the y's.
The counterfeit is a w, x, y or z respectively if both, the first, the second, or neither of these final two weighings shows inequality.

The counterfeit has been determined. The solution is "oblivious" because the exact same weighings are made regardless of results.

Now let's suppose your balance scales are unreliable. During any test sequence they may hallucinate an incorrect answer once or twice (or not at all).

(2) With two suspect coins (exactly one of which is counterfeit) there is an oblivious procedure to locate the counterfeit with five weighings (up to
two of which may report an erroneous result). Can you find such a procedure?

Weigh A vs Good three times. Weigh B vs Good twice.
If the A vs Good's all give the same result then that is the answer.
Otherwise there is an A lie, so at most one B lie.
Thus if the B results match (and the A's do not), the B result is correct. Finally if there's a mismatch on the B vs Good's then at least one was a lie, and majority rules on the A vs Good's.

...

Good Luck!
James

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From Newsgroup: rec.puzzles

On 18/07/2026 12:02, James Dow Allen wrote:

Traffic in r.p was about 20 messages per week recently, but in the last two weeks the only posts are by me. :-(

Do I have B.O. or something? (I don't think you can smell me from where you are. Could it be that today's smart-phones have secret odor sensors and are sending my stats to some place on the Dark Dark Web?)

Anyway, here are spoilers for some of the weighings solutions.
I'll insert an unrelated alt.math.rec problem as spoiler space in case anyone still wants to work on the weighings.

============ Here is unrelated puzzle as spoiler space ================

"My name is James and I'm a YouTubeaholic."
I'm not overly ashamed of this addiction. It's much less expensive than my other major addiction; and, I think, preferable to being a Facebook addict
or Xwitter critter. With Firefox frequently going into irrecoverable stall (as I experiment to find the operational maximum on the number of open tabs), I'm often presented with YouTubes I WANT to click, but avoid clicking. We know
from the Manhattan project k_eff > 1 will lead to unlimited tab explosion.

I was NOT tempted to click a recent YouTube where the title gave
BOTH a problem in geometry AND its solution.

A circle inscribed in 3-4-5 triangle has area of pi.

Whippee! I set my aging brain to work and, suspecting I had overlooked
some clever shortcut, found three equations (two linear, one quadratic)
in three unknowns. Given the answer (pi) we have a 4th equation, but
they were cumbersome enough that I hardly wanted to continue. I did though; everything canceled and r = 1 = 1 was the happy result!

Feel free to post your more clever solution and show me up.

Hmm, so suppose the triangle is ABC, with |AB|=3, |AC|=4, |BC|=5, so A is the right-angle vertex of
the triangle.

Let a = distance from A to where circle touches AB
= distance from A to where circle touches AC

[distances are the same by similar triangles etc. the centre of the circle is on the bisector of
AB,AC. This is the only bit of my answer taking any visual effort, but is clear if you draw the
diagram...]

Similarly:

Let b = distance from B to where circle touches BA
= distance from B to where circle touches BC
and
Let c = distance from C to where circle touches CA
= distance from C to where circle touches CB

So: (looking at the 3 sides of the 3,4,5 trian

a + b = 3
a + c = 4
b + c = 5

Solve. We specifically will want a, so e.g. eliminating first b then c:
c = 4-a
b+4-a = 5
b = a+1
a+a+1=3
a = 1

But a is the radius of the inscribed circle [draw own diagram!], so the area of the circle is Pi*a^2
= Pi.

Mike.

(I don't have time right now to look at the coin problem!)

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From Newsgroup: rec.puzzles


Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 18/07/2026 12:02, James Dow Allen wrote:

A circle inscribed in 3-4-5 triangle has area of pi.

...

Feel free to post your more clever solution and show me up.

Hmm, so suppose the triangle is ABC, with |AB|=3, |AC|=4, |BC|=5, so A is the right-angle vertex of
the triangle.

Let a = distance from A to where circle touches AB
= distance from A to where circle touches AC

[distances are the same by similar triangles etc. the centre of the circle is on the bisector of
AB,AC. This is the only bit of my answer taking any visual effort, but is clear if you draw the
diagram...]

Similarly:

Let b = distance from B to where circle touches BA
= distance from B to where circle touches BC
and
Let c = distance from C to where circle touches CA
= distance from C to where circle touches CB

So: (looking at the 3 sides of the 3,4,5 trian

a + b = 3
a + c = 4
b + c = 5

Solve. We specifically will want a, so e.g. eliminating first b then c:
c = 4-a
b+4-a = 5
b = a+1
a+a+1=3
a = 1

But a is the radius of the inscribed circle [draw own diagram!], so the area of the circle is Pi*a^2
= Pi.

Mike.

Nice. This isn't the first (nor the 2nd nor the 3rd) case where
I've overlooked an elegant solution based on similar or isosceles triangles.

Just now the combination of circle and right triangle triggered my memory
of a very elegant proof of Euler's Basel Identity:
https://www.youtube.com/watch?v=d-o3eB9sfls
The video's proof uses the "little-known Inverse Pythagorean Triangle".
a^-2 + b^-2 = h^-2
when a^2 + b^2 = c^2 and h is the altitude from the hypotenuse
to the ab vertex. (For example, in the 3-4-5 triangle the altitude
is 2.4 since 1/9 + 1/16 = 1/2.4^2 )
But the video's elegant proof is much MUCH more interesting than just
the Inverse Pythagorean Triangle.

Cheers,
James

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From Newsgroup: rec.puzzles

On 7/18/2026 4:02 AM, James Dow Allen wrote:

A circle inscribed in 3-4-5 triangle has area of pi.

Reminded me of the puzzle from "Mathematical Games" column:

What is the area of the largest ellipse inscribed in a 3-4-5 triangle?
--
Carl G.


--
This email has been checked for viruses by AVG antivirus software.
www.avg.com
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From Newsgroup: rec.puzzles


"Carl G." <carlgnews@microprizes.com> posted:

A circle inscribed in 3-4-5 triangle has area of pi.

Reminded me of the puzzle from "Mathematical Games" column:

What is the area of the largest ellipse inscribed in a 3-4-5 triangle?

Zl jvyq thrff vf:
Ab vafpevorq ryyvcfr jvyy unir terngre nern guna gur pvepyr vgfrys.
ORPNHFR vs gurer jrer fhpu fcrpvny ryyvcfr(f) vg jbhyq or evqvphybhfyl
fnqvfgvp gb rkcrpg n abezl gb vqragvsl vg naq pbzchgr vgf nern.


Purref,
Wnzrf
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From Newsgroup: rec.puzzles

On 7/20/2026 11:07 AM, James Dow Allen wrote:

"Carl G." <carlgnews@microprizes.com> posted:

A circle inscribed in 3-4-5 triangle has area of pi.

Reminded me of the puzzle from "Mathematical Games" column:

What is the area of the largest ellipse inscribed in a 3-4-5 triangle?

Zl jvyq thrff vf:
Ab vafpevorq ryyvcfr jvyy unir terngre nern guna gur pvepyr vgfrys.
ORPNHFR vs gurer jrer fhpu fcrpvny ryyvcfr(f) vg jbhyq or evqvphybhfyl fnqvfgvp gb rkcrpg n abezl gb vqragvsl vg naq pbzchgr vgf nern.

Purref,
Wnzrf

The largest ellipse is not a circle. The circle is the largest ellipse
for an equilateral triangle, but not a 3-4-5 triangle.

I found it easier to figure out the elipse's area than the dimensions of
the ellipse's axes or its orientation.

Hint: Gur engvb bs gur pvepyr'f nern gb gur rdhvyngreny gevnatyr'f nern
vf cv gvzrf gur fdhner ebbg bs guerr, qvivqrq ol avar.
--
Carl G.



--
This email has been checked for viruses by AVG antivirus software.
www.avg.com
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From Newsgroup: rec.puzzles


James Dow Allen <user4353@newsgrouper.org.invalid> posted:

"Carl G." <carlgnews@microprizes.com> posted:

A circle inscribed in 3-4-5 triangle has area of pi.

Reminded me of the puzzle from "Mathematical Games" column:

What is the area of the largest ellipse inscribed in a 3-4-5 triangle?

Zl jvyq thrff vf:
Ab vafpevorq ryyvcfr jvyy unir terngre nern guna gur pvepyr vgfrys.
ORPNHFR vs gurer jrer fhpu fcrpvny ryyvcfr(f) vg jbhyq or evqvphybhfyl fnqvfgvp gb rkcrpg n abezl gb vqragvsl vg naq pbzchgr vgf nern.

Purref,
Wnzrf

See https://en.wikipedia.org/wiki/Steiner_inellipse
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From Newsgroup: rec.puzzles

On Sun, 19 Jul 2026 02:43:28 +0100, Mike Terry <news.dead.person.stones@darjeeling.plus.com> wrote:

On 18/07/2026 12:02, James Dow Allen wrote:

......
A circle inscribed in 3-4-5 triangle has area of pi.

Whippee! I set my aging brain to work and, suspecting I had overlooked
some clever shortcut, found three equations (two linear, one quadratic)
in three unknowns. Given the answer (pi) we have a 4th equation, but
they were cumbersome enough that I hardly wanted to continue. I did though; >> everything canceled and r = 1 = 1 was the happy result!

Feel free to post your more clever solution and show me up.

I thought that you were joking .....

Hmm, so suppose the triangle is ABC, with |AB|=3, |AC|=4, |BC|=5, so A is the right-angle vertex of
the triangle.

Let a = distance from A to where circle touches AB
= distance from A to where circle touches AC

[distances are the same by similar triangles etc. the centre of the circle is on the bisector of
AB,AC. This is the only bit of my answer taking any visual effort, but is clear if you draw the
diagram...]

Similarly:

Let b = distance from B to where circle touches BA
= distance from B to where circle touches BC
and
Let c = distance from C to where circle touches CA
= distance from C to where circle touches CB

So: (looking at the 3 sides of the 3,4,5 trian

a + b = 3
a + c = 4
b + c = 5

Solve. We specifically will want a, so e.g. eliminating first b then c:
c = 4-a
b+4-a = 5
b = a+1
a+a+1=3
a = 1

But a is the radius of the inscribed circle [draw own diagram!], so the area of the circle is Pi*a^2
= Pi.

Actually, all once needs here is the fact that
the two tangents to a cirlce from a point without
the circle are of equal length.

Using Mike's geometry one can make the following
argument. For clarity, I will add point P, Q and R which are
where the incircle touches, respectively, the
sides BC, CA and AB.

Per Mike's definition, AR = a. Hence, BR = 3- a.
But, BR = BP. So BP = 3 - a.

By the same logic, CP = 4 - a as AQ = a.
(If required, draw the radii OP, OQ and
OR from the centre of the incircle O.
Deduce ORAQ is a square of side a.)

But BC = BP + PC. Hence,

5 = 3 - a + 4 - a

Therefore, a = 1.

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On Mon, 20 Jul 2026 19:22:27 GMT, Ilan Mayer
<user4643@newsgrouper.org.invalid> wrote:

See https://en.wikipedia.org/wiki/Steiner_inellipse

Thank you! Never heard of it till now, but
Steiner keeps crawling out of the woodwork.
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From Newsgroup: rec.puzzles


Charlie Roberts <croberts@gmail.com> posted:

On Mon, 20 Jul 2026 19:22:27 GMT, Ilan Mayer <user4643@newsgrouper.org.invalid> wrote:

See https://en.wikipedia.org/wiki/Steiner_inellipse

Thank you! Never heard of it till now, but
Steiner keeps crawling out of the woodwork.

The Wikipedia article cites Problem 98 in D||rrie's famous

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From Newsgroup: rec.puzzles


Charlie Roberts <croberts@gmail.com> posted:

On Mon, 20 Jul 2026 19:22:27 GMT, Ilan Mayer <user4643@newsgrouper.org.invalid> wrote:

See https://en.wikipedia.org/wiki/Steiner_inellipse

Thank you! Never heard of it till now, but
Steiner keeps crawling out of the woodwork.

The Wikipedia article cites Problem 98 in D||rrie's famous
*100 Great Problems of Elementary Mathematics
I think the key point is that the affine transform of any
ellipse is another ellipse (or a circle).

Tell me if I have the following correct:
[] a rigid transform is a rotation or translation or
(optionally?) reflection, or a concatenation of
such transforms.
[] a scaling transform has the form (x,y) --> (Ariax, Briay), and
yields affine transforms when concatenated with rigid transforms
[] an angle-preserving affine transform is a concatenation of
rigid transforms and dilations (scaling transforms where B=A)
[] an area-preserving affine transform is a concatenation of
rigid transforms and squeezes (scaling transforms where B=1/A)
[] a "shear" is the affine transform most difficult to think about,
but can be ignored since it can be treated as the
concatenation of (easily understood) rotations and scalings.

Steiner's result is mentioned briefly in the 2nd paragraph of
https://fabpedigree.com/james/gmat200.htm#Steiner --
a mini-bio I'm CERTAIN I read several years ago though my amnesia
is getting worse. "Constructing an ellipse" also depends on Problem 42
in D||rrie's book, and will require a special device, e.g. the
Pin-and-String method.

Most of the Problems in D||rrie's book have one (sometimes two)
associated name in the Problem's title. Unless I erred only six
mathematicians appear in the T.O.C. more than twice:
8 Steiner
6 Euler
5 Gauss
4 Newton
4 Fermat
3 Archimedes


Cheers,
James

------------------
I despise interfaces -- of which newsgouper is unfortunately an example -- which interpret keys easily misstruck by fat fingers as shortcuts
for "Post." Apologies to the newsgroup for my fat finger:

James Dow Allen <user4353@newsgrouper.org.invalid> defecated:

The Wikipedia article cites Problem 98 in D||rrie's famous

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