Advanced book on Mathematics Olympiad

Geometry and Trigonometry 613

A

M

B

O

C

x

y P

Figure 74

HenceChas coordinates( 0 , 2 cbc−a).
The slope of the lineCMisba, so the equation of this line is

y=

b

a

x+

bc

2 c−a

.

Intersecting it withAP, whose equation is

y=

b

a−c

x+

bc

c−a

,

we obtainMof coordinates( 2 cac−a, 2 c^2 bc−a). This point lies on the liney=^2 abx, so this line
might be the locus.
One should note, however, thatA=Oyields an ambiguous construction, so the
origin should be removed from the locus. On the other hand, any(x, y)on this line
yields a pointc, namely,c= 2 xax−a, except forx=a 2. Hence the locus consists of the
line of slope^2 abthrough the origin with two points removed.
(A. Myller,Geometrie Analitica ̆(Analytical Geometry), 3rd ed., Editura Didactica ̧ ̆si
Pedagogica, Bucharest, 1972) ̆

595.First, assume thatABCDis a rectangle (see Figure 75). LetHbe the intersection
point ofFGandBD. In the right trianglesABCandFBG, the segmentsBEandBH
are altitudes. Then∠ABE=∠ACBand∠BGF=∠HBC. Since∠HBC=∠ACB,
it follows that∠GBE=∠BGFandBE=GE. This implies that in the right triangle
BGF,GE=EF.
For the converse, we employ coordinates. We reformulate the problem as follows:
Given a triangleABCwithAB =BC, letBEbe the altitude fromBandOthe
midpoint of sideAC. The perpendicular fromEtoBOintersectsABatGandBCat
F. Show that if the segmentsGEandEFare equal, then the angle∠Bis right.