From Newsgroup: rec.puzzles

From 'Amusements in Mathematics' by Henry Ernest Dudeney.

Here is an easy geometrical puzzle. The crescent is formed by two
circles, and C is the centre of the larger circle. The width of the
crescent between B and D is 9 inches, and between E and F 5 inches. What
are the diameters of the two circles?

The image is available on-line at the following link:

https://en.wikisource.org/wiki/Page:Amusements_in_mathematics.djvu/64

For anyone who doesn't use the Worldwide web, the set-up is described below:

[Draw a large circle and mark the centre 'C'. Draw a horizontal line
through C meeting the circle at points 'B' (left) and 'A' (right). Draw
a vertical line through C meeting the circle at point 'F' (top) and an unmarked point at the bottom. Draw another, somewhat smaller circle,
with its centre on the line BA and right extremity passing through point
A. Mark the point where this smaller circle crosses the vertical centre
line 'E'. You should now have two circles that touch at there right
extremity and form a crescent shape between them. The distance FE is 5"
and the distance BD is 9".]
--
David Entwistle

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From Newsgroup: rec.puzzles

On 30/08/2025 09:01, David Entwistle wrote:

[Draw a large circle and mark the centre 'C'. Draw a horizontal line
through C meeting the circle at points 'B' (left) and 'A' (right). Draw
a vertical line through C meeting the circle at point 'F' (top) and an unmarked point at the bottom. Draw another, somewhat smaller circle,
with its centre on the line BA and right extremity passing through point
A. Mark the point where this smaller circle crosses the vertical centre
line 'E'. You should now have two circles that touch at there right extremity and form a crescent shape between them. The distance FE is 5"
and the distance BD is 9".]

['D' is the point where the smaller circle crosses the horizontal
centreline.]
--
David Entwistle
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From Newsgroup: rec.puzzles

In article <108ub56$2frd9$2@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Here is an easy geometrical puzzle. The crescent is formed by two
circles, and C is the centre of the larger circle. The width of the
crescent between B and D is 9 inches, and between E and F 5 inches. What
are the diameters of the two circles?

And what if the distance between E and F is 4 inches?

The image is available on-line at the following link:

https://en.wikisource.org/wiki/Page:Amusements_in_mathematics.djvu/64

-- Richard
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From Newsgroup: rec.puzzles

On Sat, 30 Aug 2025 09:04:59 +0100, David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

On 30/08/2025 09:01, David Entwistle wrote:

[Draw a large circle and mark the centre 'C'. Draw a horizontal line
through C meeting the circle at points 'B' (left) and 'A' (right). Draw
a vertical line through C meeting the circle at point 'F' (top) and an
unmarked point at the bottom. Draw another, somewhat smaller circle,
with its centre on the line BA and right extremity passing through point
A. Mark the point where this smaller circle crosses the vertical centre
line 'E'. You should now have two circles that touch at there right
extremity and form a crescent shape between them. The distance FE is 5"
and the distance BD is 9".]

['D' is the point where the smaller circle crosses the horizontal >centreline.]

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Let us name the point where the vertical centre line intersects
the smaller circle below BA (i.e. the reflection of E by BA) G.
Also, let the radius of the larger circle be R.

Then EG and DA are two chords of the smaller circle that
intersect at C.

Hence EC.CG = DC.CA

But, EC = CG = R - 5
DC = R - 9

Hence we get (R - 5)^2 = R(R - 9)

which gives R = 25.

The radius of the smaller cirlce is

AD/2 = (R + R - 9)/2 = 41/2
--
This email has been checked for viruses by Avast antivirus software. www.avast.com
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From Newsgroup: rec.puzzles

In article <108ub56$2frd9$2@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote: >https://en.wikisource.org/wiki/Page:Amusements_in_mathematics.djvu/6

Another solution:

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Call the radii of the larger and smaller circles R and r. Let H be
the centre of the smaller circle.

EH = r = R - 9/2
EC = R - 5
CH = 9/2

By Pythagoras, EH^2 = EC^2 + CH^2:

(R - 9/2)^2 = (R - 5)^2 + (9/2)^2

Expanding,
R^2 - 9R + (9/2)^2 = R^2 - 10R + 25 + (9/2)^2

Cancelling,
-9R = -10R + 25

R = 25
r = 41/2

-- Richard
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From Newsgroup: rec.puzzles

In article <l6m6bk92h5hs1q4vkrckniundlev0f20nl@4ax.com>,
Charlie Roberts <croberts@gmail.com> wrote:

On Sat, 30 Aug 2025 09:04:59 +0100, David Entwistle ><qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

On 30/08/2025 09:01, David Entwistle wrote:

[Draw a large circle and mark the centre 'C'. Draw a horizontal line
through C meeting the circle at points 'B' (left) and 'A' (right). Draw >>> a vertical line through C meeting the circle at point 'F' (top) and an
unmarked point at the bottom. Draw another, somewhat smaller circle,
with its centre on the line BA and right extremity passing through point >>> A. Mark the point where this smaller circle crosses the vertical centre >>> line 'E'. You should now have two circles that touch at there right
extremity and form a crescent shape between them. The distance FE is 5" >>> and the distance BD is 9".]

['D' is the point where the smaller circle crosses the horizontal >>centreline.]

SPOILER BELOW
SPOILER BELOW
SPOILER BELOW
SPOILER BELOW

SPOILER BELOW
SPOILER BELOW
SPOILER BELOW
SPOILER BELOW

SPOILER BELOW
SPOILER BELOW
SPOILER BELOW
SPOILER BELOW

Let us name the point where the vertical centre line intersects
the smaller circle below BA (i.e. the reflection of E by BA) G.
Also, let the radius of the larger circle be R.

Then EG and DA are two chords of the smaller circle that
intersect at C.

Hence EC.CG = DC.CA

If we happen to have forgotten the intersecting chords theorem, we
can obtain EC/CA = DC/CE by observing that AED is a right angle
(angle in a semicircle) so DEC and ECA are similar.

But, EC = CG = R - 5
DC = R - 9

Hence we get (R - 5)^2 = R(R - 9)

which gives R = 25.

The radius of the smaller cirlce is

AD/2 = (R + R - 9)/2 = 41/2

--
This email has been checked for viruses by Avast antivirus software. >www.avast.com

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From Newsgroup: rec.puzzles

On Sat, 30 Aug 2025 10:58:31 -0000 (UTC), richard@cogsci.ed.ac.uk
(Richard Tobin) wrote:

In article <108ub56$2frd9$2@dont-email.me>,

And what if the distance between E and F is 4 inches?

Then BD < 9!

No, I do not mean to be facetious. The aim of this posting is to see
if that irritating thing Avast adds to email and Usenet posts no
longer occur. I keep turning it off, but the gremlins seem to like
turning it on again.

OTOH, you could ask was BD would be if EF = 4. A problem in
itself.

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From Newsgroup: rec.puzzles

On 30/08/2025 22:30, Richard Tobin wrote:

If we happen to have forgotten the intersecting chords theorem, we
can obtain EC/CA = DC/CE by observing that AED is a right angle
(angle in a semicircle) so DEC and ECA are similar.

If you are wondering what the intersecting chords theorem is, or are struggling with the problem, or don't know where to start (I fit into
one of those groups), NRICH have some similar problems with hints and guidance.

https://nrich.maths.org/problems/partly-circles?tab=overview

NRICH have some very nice information on many topics.
--
David Entwistle
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From Newsgroup: rec.puzzles

In article <108ub56$2frd9$2@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

https://en.wikisource.org/wiki/Page:Amusements_in_mathematics.djvu/64

Dudeney's own answer is rather cryptic. He refers to one length being
a mean proportional between two others (by which he means that it is
the geometric mean of them), but doesn't explain why that is the case.

Presumably he could expect his readers to be familiar with the
altitude theorem:

https://en.wikipedia.org/wiki/Geometric_mean_theorem

which derives it using similar triangles and from the intersecting
chords theorem as described elsewhere in this thread.

I don't recall being taught it explicitly at school.

-- Richard
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From Newsgroup: rec.puzzles

On Sun, 31 Aug 2025 12:27:33 -0000 (UTC), richard@cogsci.ed.ac.uk
(Richard Tobin) wrote:

Dudeney's own answer is rather cryptic. He refers to one length being
a mean proportional between two others (by which he means that it is
the geometric mean of them), but doesn't explain why that is the case.

Presumably he could expect his readers to be familiar with the
altitude theorem:

https://en.wikipedia.org/wiki/Geometric_mean_theorem

which derives it using similar triangles and from the intersecting
chords theorem as described elsewhere in this thread.

I don't recall being taught it explicitly at school.

I remember this being taught this, though I cannot recollect
whether is was left as "an exercise to the reader" or was an
exercise. Our text was "A School Geometry" by Hall and
Stevens. One version can be found at

https://archive.org/details/schoolgeometry00hall/page/n3/mode/2up

I see that Sricbd has the 1910 version of all three volumes.

(Such Edwardian texts were still the norm in many schools
in India well into the late 1970s. Local versions -- essentially
pirated and often verbatim -- replaced them.)



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