Show that for any triangle it is always possible to construct 3
touching circles with centres at the vertices. Is it possible to
construct touching circles centred at the vertices of any polygon?
Find the vertices of a pentagon given the midpoints of its sides.
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.
Two solutions offered by Andrei of School 205, Bucharest. Many
I solved the problem in two ways:
I find from reasons of symmetry the equality of areas
Without loss of generality, I suppose x < l/2, and y <
I start from the observation that there are from the beginning
two congruent triangles, the one coloured in purple, and the other
with purple lines.
I look now to the trapezium coloured continuously in red, and I
try to find another one, congruent with this one. For this, I trace
a line AB parallel to the base of the square, at 2y. It is drawn in
I see the equality of the areas of the two trapeziums the one in
continuous red, the other with red lines.
The line drawn determines another 4 trapeziums, with equal areas
two by two, and that could be identified from the colour: green and
Now, I remained with all the area above the line AB just drawn.
To find equal areas, I drew another line, parallel with the
vertical side of the square, at 2x.
The equal areas found are again with the same colour, line and
Now, the part of the pizza with lines has the same area with the
part of the pizza with continuous colours, and it is easy to join
the parts that have been determined by the new lines.
I calculated the areas of all figures (triangles and rectangular
trapeziums) delimitated by the sides of the pizza and by the
I noted all the points from the figure (see chart). Now, let the
length of DI be x and the length of DK be y; let the side of the
square be 1 unit.
A LMK = x 2 /2
A KJM = x 2 /2
A DJMI = xy - x 2 /2
A HCGM = 1y - xy - y 2 /2
A MFBG = (1 2 - y 2 )/2 - 1x +
A LAEM = 1x - xy - x 2 /2
A EMF = 1 2 /2 - 1y + y 2 /2
Now, look what areas to add to obtain 1 2 /2.
First I take triangle EMF, whose area begins with 1 2 /2. Because 1y is subtracted from 1 2 /2, I must look for an area that contains (+ 1y).
The only area containing such a number is MHCG. I write the
obtained area up to now:
AEMF + AHCGM = 1 2 /2 - 1y + y
2 /2 + 1y - xy - y 2 /2 = 1 2 /2 - xy
Now, I must add something containing (+xy). I add LKM and DJMI,
obtaining 1 2 /2. For the remaining
areas the sum is naturally 1 2 /2.
So, I added the areas in the following manner:
ALKM + ADJMI + AMHCG + AEMF and
AKJM + AMIH + AMFBG + ALAEM.
These two kind of pieces are marked differently on the