Re: geometry with difficult...



On 27 Sty, 20:25, "Ignacio Larrosa Cañestro"
<ilarrosaQUITARMAYUSCU...@xxxxxxxxxxx> wrote:
En el mensaje:1168707273.354490.250...@xxxxxxxxxxxxxxxxxxxxxxxxxxx,
Zbigniew Karno <zbigniew.ka...@xxxxx> escribió:

johnjo napisal(a):
Zbigniew Karno wrote:
( I apologise for not snipping.)

My solution is purely geometric
- uses only compasses and ruler.I think that your solution can be simplifiea a little.





Because the solution follows from a geometric
construction, first I would like to establish
notation, which is a little different than yours,
but it is very symmetric.

For i = 1,2 let A_iB_iC_i denotes a triangle,
with sides of length  a_i =  B_iC_i , b_i = A_iC_i
 and c_i = A_iB_i .
Let O_i denotes the circumcentre and R_i denotes
the radius of circumcircle S_i of the triangle
A_iB_iC_i .

Your problem can be reformulated as follows:

Problem.
  Suppose that A_2 is the midpoint of the side B_1C_1
and A_1 is the midpoint of the side B_2C_2.
In particular, (c_1)/2 = B_1A_2 = A_2C_1 and
(c_2)/2 = B_2A_1 = A_1C_2 in length, moreoverIn particular, (a_1)/2 = B_1A_2 = A_2C_1 and
(a_2)/2 = B_2A_1 = A_1C_2 in length, moreover





the segment A_1A_2 is a common median of both
triangles.
Suppose also that the line B_1C_1 is a bisector of the
angle <A_2 =  <B_2A_2C_2 of the triangle A_2B_2C_2
and the line B_2C_2 is a bisector of the angle
<A_1 =  <B_1A_1C_1 of the triangle A_1B_1C_1 .
   Then c_1 + b_1 = c_2 + b_2.

Solution with constructions:

Step 1. Triangle A_1B_1C_1 .
  Consider arbitrary triangle A_1B_1C_1, but
with the property that b_1 is not equal c_1
(otherwise, the triangle A_2B_2C_2,
constructed in Step 2, will be degenerate).
It is easy to construct the circumcircle S_1
of A_1B_1C_1.
In particular, the mid-perpendicular of the side
B_1C_1 contains the circumcentre O_1 of S_1
and intersects S_1 at two points, say F_1 and G_1,
so that the segment F_1A_1 bisects the angle <A_1
(it is right because the segments B_1F_1 and C_1F_1
- chords of S_1 - have the same length).
Observe that the points G_1 and A_1 are different,
because b_1 is not equal c_1. Note also that the
segment G_1F_1 is a diameter of S_1.

Step 2. Triangle A_2B_2C_2 .
  Note that point A_2 is determined, it is the
midpoint of the segment  B_1C_1. It remains
to determine B_2 and C_2 .
First, I would like to mark out the point F_2 and
the circumcentre O_2 of S_2 and then to determine
the radius R_2 and where S_2 is.

Observe that the line B_1C_1 should be a bisector
of the angle <A_2  of the triangle A_2B_2C_2, and
A_1 should be a midpoint of B_2C_2. Moreover,
since the line B_2C_2 should be a bisector of the
angle <A_1 of the triangle A_1B_1C_1, it follows
that the line perpendicular to the line F_1A_1 at the
point A_1 will be the mid-perpendicular of the side
B_2C_2.
Observe also that the triangle F_1A_1G_1 is right,
therefore the line G_1A_1 is perpendicular to the
line F_1A_1 at A_1, so this line should serves as the
mid-perpendicular of the side B_2C_2 of  desired
triangle A_2B_2C_2 and it should contains the
circumcentre O_2 of S_2.
Thus the lines G_1A_1 and B_1C_1 meets at
a point, say F_2.

Since the segment A_2F_1 should be a chord of

I think this should be A_2F_2

the circumcircle S_2, it follows that its
mid-perpendicular should contains the circumcentre
O_2 of S_2, but this line meets the line G_1A_1.
Hence this intersection point is exactly the
circumcentre O_2 of  S_2.
Now the length of the segment A_2O_2 equals R_2.
Moreover, since the lines F_1G_1 and A_2F_2 are
perpendicular at A_2, it follows that the length of
the segment O_2G_1 is also equals R_2.
In this way, it is possible to determine exactly, where
the circumcircle S_2 lives.

Note that the line O_2F_2 (= the line G_1F_2) meets
S_2 at F_2 and G_2 = G_1. In particular, the segment
G_2F_2 is a diameter of S_2.

Now it is a proper moment to determine points
B_2 and C_2. These are intersection points of S_2
and the line F_1A_1. The triangle A_2B_2C_2
has all desired properties.

Step 3. The circumcircle of the
                quadrilateral B_1B_2C_1C_2.
   Consider the point G = G_1 =G_2. Note that
the intersection of S_1 and S_2 consists of two
points, namely G and other point, say H.
The line GH is the radical axis of the circles
S_1 and S_2. From this and from the facts
that GF_1 is the diameter of S_1 and GF_2
is the diameter of S_2, it follows that the
orthocenter, say E, of the triangle F_1GF_2
is exactly the common point of intersection
of three lines B_1F_2, C_2F_1 and GH.I would substitute the next:
===========



From this it follows that:
 (*) The point E is the radical point of
three circles S_1, S_2 and S, where S
denotes the circle for which the
following holds:
  the line B_1C_1 is the radical line
  of S_1 and S
 or
  the line B_2C_2 is the radical line
  of S_2 and S.
(In fact such circle S is unique.)

Observe that the segments B_1G and C_1G
have the same length. The same holds for
the segments B_2G and C_2G. However,
the property (*) implies that all these four
segments have the same length, say R.
Let S be the circle with the centre G and
with the radius R. It follows that S contains
four point B_1, B_2, C_1 and C_2 .
It means that S is the circumcircle of the
quadrilateral B_1B_2C_1C_2 and the
segments B_1C_1 and B_2C_2 are its
diagonals.

Im with this so far===================
By:

Let S be the circle with the centre G passing throw B1 and C1. The radical
axis of S and S1 is then the line B1C1. Then E, intesection of B1C1 and GH,
is the radical centre of S, S1 and S2. This implies that the radical axis of
S and S2 is the line throw E perpendicular to GO2; i.e., the line B2C2, and
then B2 and C2 are in the circle S.





Step 4. Equality c_1 + b_1 = c_2 + b_2.
 Now consider two intersections:
  the circle S and the line A_2C_2
  consisting of the points C_2 and B_2'
and
  the circle S and the line B_1A_1
  consisting of the points B_1 and C_1' .

Since B_1C_1 is the bisector of the angle
<A_2 and B_2C_2 is the bisector of the
angle <A_1, it is easy to see that
the segments A_2B_2 and A_2B_2'
have the same length, similarly
the segments A_1C_1 and A_1C_1'
have the same length.
Therefore the length of B_1C_1' equals
c_1 + b_1 and the length of C_2B_2'
equals c_2 + b_2 .To finish from here, observe that arcs of S B1B2' and B2C1 are symmetrics
with respect to the line F1G, and then B1B2' = C1B2 . At the same way, the
arcs C2C1' and C1B2 are symmetrics respect to the line F2G. Then segments
B1B2' = C2C1'

Then, considering the triangles inscribed in S B1C2B2' and  B1C2C1' we
deduce that B1C1' = C2B2'  and then finally c1 + b1 = c2 + b2.

Very nice reasoning. These arguments
considerably simplifying the proof.



Also I think would be more simetrical interchange the names of B2 and C2, as
B2' and C2' also.

--
Best regards,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCU...@xxxxxxxxxxx





 Now observe (very carefully)   that
the segment B_1C_1', and similarly
the segment  C_2B_2' are in fact the
"hidden" diagonals of  the
 quadrilateral B_1B_2C_1C_2.

aha, now my geometry knowledge fails me. What is a "hidden" diagonal.
They seem to stick out of the quadrilateral

Hence these must have the same
length, and therefore

That would do it, but why. Is there a theorem I dont know about
diagonals of same area
cyclic quadrilaterals? Ptolemy is much more complicated than that for
instance.

               c_1 + b_1 = c_2 + b_2.

To clearness the situation note that
the segment B_1C_1', with B_2C_2,
is the diagonal of the
 quadrilateral B_1B_2C_1'C_2,
and similarly
the segment C_2B_2', with B_1C_1,
is the diagonal of the
 quadrilateral B_1B_2'C_1C_2.
These quadrilaterals have the same
areas as the quadrilateral
                    B_1B_2C_1C_2
(It is easy to observe: in the first case
cut along B_2C_2 and then glue after
rotation round about the line GF_2
- A_1 is a midpoint of  the segment
B_2C_2, similarly in the second case).

I agree with this - this is because (in the first case) C_1C_1' is
parallel to B_2C_2 since
both are perpendicular to A_1F_2
Hence area triangle B_2C_1C_2 = area triangle B_2C_1'C_2

This finish my solution of the problem.

Are the constructions and the reasoning
clear ?

A little more explanation would help me.
Oh and a diagram!
Although I now have a multicoloured draft on the back of a
spread***.

Z. Karno

BTW I tried to do this with complex numbers/vectors but the trick did
not emerge.
Best wishes
JJ

Thanks for interesting comments.

Intuitions with "hidden" diagonals
in a cyclic quadrilateral are almost
clear. If ABCD is such quadrilateral,
with side lengths a = AB, b = BC, c = CD
and d = DA, then the segments AC and BD
are its "visible" diagonals, with lengths
u = AC and v = BD.
A "hidden" diagonal of ABCD is one of
new "visible" diagonals a quadrilateral
which is created by the operation of cutting
along fixed diagonal, AC or BD, of the
quadrilateral ABCD, and gluing after
pi-rotation of one triangle-piece without
moving another triangle-piece. Obviously
such a "hidden" diagonal stick out of ABCD.
Moreover, there are at most four "hidden"
diagonals of ABCD, however the lengths of
these are equal each other, say w.
Then the area K of the quadrilateral ABCD
is given by formula K = (uvw)/(4R), where
R is the circumradius of ABCD. This- Ukryj cytowany tekst -- Pokaż cytowany tekst -- Ukryj cytowany tekst -- Pokaż cytowany tekst -- Ukryj cytowany tekst -- Pokaż cytowany tekst -- Ukryj cytowany tekst -- Pokaż cytowany tekst -- Ukryj cytowany tekst -- Pokaż cytowany tekst -...

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Ignacio, thank you for interesting remarks.
Now the proof of Mina’s problem is really
elementary and easy.

Best regards,
Z. Karno


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