Little Geometry Problem

[Bill Taylor]

This comes off some puzzle website, I was told, didn't hear which one.

------------------
A circle has a square ABCD inscribed in it.

Between side AB and its neighboring arc, the largest possible square
is inscribed, (thus symmetrically), with one edge along AB and the opposite
vertices on the circle.

Prove this smaller square has edge length exactly one-fifth of that of ABCD.
------------------

It's easy enough to prove by setting up an appropriate Pythagorean quadratic,
then solving it. But the result is so neat that I feel there ought to be
a purely geometric way of proving it. But I can't find one.

Any help??

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Bill Taylor W.Ta...@math.canterbury.ac.nz
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I don't live on the edge, but sometimes I go there to visit.
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[Ignacio Larrosa Cañestro]

Bill Taylor <mat...@math.canterbury.ac.nz> escribió en el mensaje de
noticias 9rnqfo$hqp$1...@cantuc.canterbury.ac.nz...

> This comes off some puzzle website, I was told, didn't hear which one.
>
>
> ------------------
> A circle has a square ABCD inscribed in it.
>
> Between side AB and its neighboring arc, the largest possible square
> is inscribed, (thus symmetrically), with one edge along AB and the
opposite
> vertices on the circle.
>
> Prove this smaller square has edge length exactly one-fifth of that of
ABCD.
> ------------------

Let M the middpoint of side AB. Trace the straight line trhouhgt C and M.
The other point in that line intersect the circle c, let it E, is a vertix
of small square. Then, you consider the power?? of M with respect to circle:

Pot(M, c)=AM*MB=CM*ME

If the side of the big square is d, then AM=MB=d/2,
CM=sqrt(d^2+(d/2)^2)=d*sqrt(5)/2 and

ME=(d^2/4)/(d*sqrt(5)/2)=(d*sqrt(5)/2)/5=CM/5

Then also the sides are in the same ratio.

--
Saludos,

Ignacio Larrosa Cañestro
A Coruña (España)
ignacio...@eresmas.net
ICQ 94732648

[Rainer Rosenthal]

Bill Taylor <mat...@math.canterbury.ac.nz>

> A circle has a square ABCD inscribed in it.
>
> Between side AB and its neighboring arc, the largest possible

> square is inscribed, ... with one edge along AB and the

> opposite vertices on the circle.
>
> Prove this smaller square has edge length exactly one-fifth of that
> of ABCD.
>

> It's easy enough to prove by setting up an appropriate Pythagorean
> quadratic, then solving it. But the result is so neat that I feel there
> ought to be a purely geometric way of proving it. But I can't find one.

Hello Bill,

when we draw a square ABCD with sidelength 5 and attach
the 5 little squares with sidelength 1 symmetrically as given
by the problem, we get a "squary star" with outer vertices
A, 1, 2, B, 3, 4, C, 5, 6, D, 7, 8 (in this order).
Let M be the middle point of the square.

By symmetry all the distances [M,1] ... [M,8] are equal to, say, r.
And all the distances [M,A],...[M,D] are equal to sqrt(2)*5/2.

We are ready with the geometrical proof, if we can show, that

r = sqrt(2)*5/2 (*)

Consider the triangle 145. It has a right angle at point 4. We have
distances [1,4] = 4*sqrt(2) and [4,5] = 3*sqrt(2).
So this is the famous pythagorean triangle with hypotenuse of
length [1,5] = 5*sqrt(2). Since [1,5] = [M,1]+[M,5] = 2*r, we have
the desired result (*). Q.E.D.

Thanks for the nice problem
Rainer Rosenthal
r.ros...@web.de
-
I shall call this proof "The Brain of Pooh"
[ A.A. Milne, Winnie-the-Pooh
From Chapter "Surrounded by water".
(Slightly modified) ]

[Rainer Rosenthal]

Bill Taylor <mat...@math.canterbury.ac.nz>

> A circle has a square ABCD inscribed in it.
>
> Between side AB and its neighboring arc, the largest possible

> square is inscribed, ... with one edge along AB and the

> opposite vertices on the circle.
>
> Prove this smaller square has edge length exactly one-fifth of that
> of ABCD.
>

> It's easy enough to prove by setting up an appropriate Pythagorean
> quadratic, then solving it. But the result is so neat that I feel there
> ought to be a purely geometric way of proving it. But I can't find one.

Hello Bill,

here is a second version of a proof for this funny puzzle.
It's (!aha,aha!) even more geometric than its (!yeah!)
predecessor and so will suit you even better.

Again when we draw a square ABCD with sidelength 5 and

attach the 5 little squares with sidelength 1 symmetrically
as given by the problem, we get a "squary star" with outer
vertices A, 1, 2, B, 3, 4, C, 5, 6, D, 7, 8 (in this order).
Let M be the middle point of the square.

We circumscribe another square EFGH, with E near A in the
lower left corner. This square has the points A,1,B,3,C,5,D,7
in common with the squary star.
Wherever you look, you will find lots of similar triangles with
right angles and sides of ratio 1:2.
This shows pretty quick, that the line thru the middle of C and 5
and the middle of A and 1 is perpendicular and goes thru M.
The line perpendicular thru the middle of 5 and 6 and the middle
of 1 and 2 goes thru M anyway.
So M ist the circumcenter of the triangle A12 and of C56 and by
symmetry of the hole squary star.

Thanks for the nice problem
Rainer Rosenthal
r.ros...@web.de
-

I shall call this proof "The Second Brain of Pooh"

[ A.A. Milne, Winnie-the-Pooh
From Chapter "Surrounded by water".

(Strongly modified) ]