What happens to the perimeter of triangle ABC as the two smaller
circles change size and roll around inside the bigger circle?
Two semicircle sit on the diameter of a semicircle centre O of
twice their radius. Lines through O divide the perimeter into two
parts. What can you say about the lengths of these two parts?
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?
Many people noticed that however you change
the shape of the quadrilateral ABCD the inner quadrilateral is
always a parallelogram but no-one gave a proof of this fact or
found its area. You may like to try to do it with the clue that
joining the midpoints of the sides of a triangle gives two
triangles, one of which is an enlargement of the other, so for
example look at triangles ASP and ADB.
This Tough Nut was cracked early in July. Two
good solutions arrived almost simultaneously so we give them
The easiest method uses the fact that the line
joining the midpoints of the sides of a triangle is parallel to the
third side and half its length. In the diagram, consider the
enlargement centre A, scale factor 2. Triangles ASP and ADB are
similar, the lengths of the sides are in the ratio 1:2 and hence
their areas are in the ratio 1:4.
The following solution comes from Alexander of
Shevah-Mofet, Tel Aviv, Israel.
PQRS is always a parallelogram and it is very easy to
[Notation used : S(...) - area of something.
a||b - if a is parallel to b, same for AB||CD.]
Since AP= ½AB and AS= ½AD therefore SP is
parallel to DB.
Using the same reasons : RQ||DB and therefore RQ||SP.
By the same argument : PQ||SR so PQRS is a parallelogram.
The area of the parallelogram PQRS is half the area of the
quadrilateral ABCD, and here's the proof.
1. A convex quadrilateral ABCD.
2. S is the midpoint of AD.
3. P is the midpoint of AB.
4. Q is the midpoint of BC.
5. R is the midpoint of DC.
7.S(ASP) = ½ sp sin(DAB).
8.S(ADB) = ½ (2 s )(2 p )sin(DAB) =2
9.S(ASP) = ¼ S(ADB) (from 7 and 8).
10.For the same reason as 7,8,9 :
10.1 S(BPQ) = ¼ S(BAC).
10.2 S(CQR) = ¼ S(CBD).
10.3 S(DRS) = ¼ S(DCA).
13. 2*S(ABCD)=S(ABC)+S(ACD)+S(ABD)+S(DBC) (from 11 and 12).
15. S(SPQR)=S(ABCD) - ¼
(from 9,10.1,10.2,10.3 and 14).
16. S(SPQR)=S(ABCD) - ½ S(ABCD) (from 13 and 15).
17. S(SPQR) = ½ S(ABCD) (from 16).
w.w.w.l.f (what we were looking for)
The other solution, from Joel of Singapore,
came complete with diagram and a different proof that area PQRS is
½ area ABCD.
Swing triangles a and b around so that PA coincides with PB and
CQ with CB, and so that SPS 1 and RQR 1 are
straight lines. Now DS = BS 1 and DR = BR 1 .
The angles of quadrilateral ABCD add up to 360 and they fit around the
point B when triangle d slides to fit into BS 1 R
1 to form a quadrilateral. We can now show that angle S
1 SR + angle SS 1 R 1 = 180
hence that SR || S 1 R 1 .
Angle ASP = angle PS 1 B.
Angle DSR = angle BS 1 R 1 .
Angle ASP + angle DSR + angle PSR = 180 (angles on a straight
Hence angle PS 1 B + angle BS 1 R
1 + angle PSR = 180 .
As SR and S 1 R 1 are equal and parallel
it follows that SS 1 R 1 R is a
parallelogram. Moreover SP = PS 1 and RQ = QR
1 so SPQR and PS 1 R 1 Q are
congruent parallelograms . Hence the area of PQRS is half the area