Re: geometry with difficult...


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

My solution is purely geometric
- uses only compasses and ruler.

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, moreover

I think that should be (a_1)/2 and (a_2)/2

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.

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



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 .

yep


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

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