Advanced book on Mathematics Olympiad

4.1 Geometry 229

We will show that the only possible configuration is that in Figure 32. Consider a
triangle that maximizes the area (such a triangle exists since the vertices vary on compact
sets and the area depends continuously on the vertices). The vertices of this triangle must
lie on the half-circle. IfBlies betweenAandC, thenAandCmust be the endpoints
of the diameter. Indeed, if sayCis not an endpoint, then by moving it toward the closer
endpoint of the diameter we increase bothACand the angle∠BAC; hence we increase

the area. Finally, among all triangles inscribed in a semicircle


AC, the isosceles right
triangle has maximal altitude, hence also maximal area. This triangle has area 1, and the
claim is proved.

A

B

C

Figure 32

Returning to the problem, let us note that since the two triangles in question are
convex sets, they can be separated by a line. That line cuts the disk into two regions, and
one of them, containing one of the triangles, is included in a half-disk. By what we just
proved, this region must itself be a half-disk. The only possible configuration consists of
two isosceles triangles sharing the hypotenuse. 

The next problem was published by the first author in theMathematics Magazine.

Example.LetABCbe a right triangle (∠A= 90 ◦). On the hypotenuseBCconstruct
in the exterior the equilateral triangleBCD. Prove that the lengths of the segmentsAB,
AC, andADcannot all be rational.

Solution.We will find a relation betweenAB,AC, andADby placing them in a triangle
and using the law of cosines. For this, construct the equilateral triangleACEin the
exterior ofABC(Figure 33). We claim thatBE=AD. This is a corollary of Napoleon’s
problem, and can be proved in the following way. LetMbe the intersection of the
circumcircles ofBCDandACE. Then∠AMC= 120 ◦and∠DMC= 60 ◦; hence
M∈AD. Similarly,M ∈BE. Ptolemy’s theorem applied to quadrilateralsAMCE
andBMCDshows thatME=AM+CMandMD=BM+CM; henceAD=
AM+BM+CM=BE.
Applying the law of cosines in triangleABE, we obtainBE^2 =AB^2 +AE^2 +AB·
AE

√

3, and sinceBE=ADandAE=AC, it follows that

AD^2 =AB^2 +AC^2 +AB·AC

√

3.