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Stage: 4 Challenge Level: Challenge Level:1

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.

Diagram of the problem

This Tough Nut was cracked early in July. Two good solutions arrived almost simultaneously so we give them both.

A triangle with a line joining the midpoints of the sides

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 prove.
[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.

Given:
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.
Find: S(SPQR)/S(ABCD).
Search:

6. Let's mark AP=PB with p ,
BQ=QC with q ,
CR=RD with r ,
and DS=SA with s .

7.S(ASP) = ½ sp sin(DAB).
8.S(ADB) = ½ (2 s )(2 p )sin(DAB) =2 sp sin(DAB).
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).
11. S(ABCD)=S(ABC)+S(ACD).
12. S(ABCD)=S(ABD)+S(DBC).
13. 2*S(ABCD)=S(ABC)+S(ACD)+S(ABD)+S(DBC) (from 11 and 12).
14. S(SPQR)=S(ABCD)-(S(SAP)+S(BPQ)+S(CQR)+S(DRS)).
15. S(SPQR)=S(ABCD) - ¼ (S(ABC)+S(ACD)+S(ABD)+S(DBC))
(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).
18. S(SPQR)/S(ABCD)=½.
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  degrees 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  degrees and 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  degrees (angles on a straight line).
Hence angle PS 1 B + angle BS 1 R 1 + angle PSR = 180  degrees.

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 of ABCD.

Diagram of Joel Tay's solution