Everybody knows that in any triangle one can inscribe a circle whose center is at the intersection point of the angles’ bisectors.
What can we say about a quadrilateral? If there exists a circle touching all the quadrilateral’s sides, then its center is equidistant from them, hence it lies on all four angle bisectors. We deduce that a necessary and sufficient condition for the existence of an inscribed circle is that the quadrilateral’s angle bisectors are concurrent (in fact, this works for arbitrary convex polygons).
This does not happen in every quadrilateral. We can always draw a circle tangent to three of the four sides (its center being the point of intersection of two of the bisectors) (Fig.2.30).
If three of the four angle bisectors meet at one point, it is easy to see that the fourth one will also pass through that point and that a circle can be inscribed in the quadrilateral.
Another necessary and sufficient condition for the existence of an inscribed circle in a quadrilateral is given by the following theorem, due to Pithot.
Theorem LetABCD be a convex quadrilateral. There exists a circle inscribed in ABCDif and only if:
AB+CD=AD+BC.
Proof Suppose there exists a circle inscribed in the quadrilateral, touching the sides AB, BC, CDandDAat the pointsK, L, M, N, respectively. Then, since the tan-
2.7 Quadrilaterals with an Inscribed Circle 63 Fig. 2.30
Fig. 2.31
gents from a point to a circle have equal lengths, we have AK =AN, BK = BL, CM=CLandDM=DN (Fig.2.31).
If we add up these equalities, we obtain the desired result.
Conversely, supposeAB+CD=AD+BC. Draw a circle tangent toAB, BC andCD. If the circle is not tangent toAD, draw fromAthe tangent to the circle and letEbe the point of intersection withCD. Suppose, for instance, thatElies in the interior ofCD(Fig.2.32).
Since the circle is inscribed in the quadrilateralABCE, we haveAB+CE= AE+BC. On the other hand, from the hypothesis, we haveAB+CD=AD+BC, orAB+CE+ED=AD+BC. From these, we deriveED+AE=AD, which is impossible. It follows that the circle is also tangent toAD, hence it is inscribed inABCD.
IfElies outside the line segmentCD, the proof is almost identical.
Fig. 2.33
Another nice proof of the converse is the following: ifAB=ADthenBC=CD and the conclusion is immediate. Suppose, with no loss of generality, thatAB <
AD, and letX∈ADbe such thatAB=AX. LetY ∈CDbe such thatDX=DY. SinceAB+CD=AD+BC, it follows thatCY=BC(Fig.2.33).
Thus, the trianglesABX, DXY and CY B are isosceles, so the perpendicular bisectors of the sidesBX, XYandY Bof triangleBXY are also the angle bisectors of∠A,∠D, and∠C of the quadrilateral. Since the perpendicular bisectors of the sides of a triangle are concurrent, we obtain the desired conclusion.
Problem 2.60 Prove that if in the quadrilateralABCD is inscribed a circle with centerO, then the sum of the angles∠AOBand∠CODequals 180◦.
Problem 2.61 LetABCDbe a quadrilateral with an inscribed circle. Prove that the circles inscribed in trianglesABCandADCare tangent to each other.
Problem 2.62 LetABCDbe a convex quadrilateral. Suppose that the linesABand CDintersect atEand the linesADandBCintersect atF, such that the pointsE andF lie on opposite sides of the lineAC. Prove that the following statements are equivalent:
2.7 Quadrilaterals with an Inscribed Circle 65 Fig. 2.34
(i) a circle is inscribed inABCD;
(ii) BE+BF=DE+DF; (iii) AE−AF=CE−CF.
Problem 2.63 Let ABCD be a convex quadrilateral. Suppose that the lines AB andCDintersect atEand the linesADandBCintersect atF. LetMandN be two arbitrary points on the line segmentsAB andBC, respectively. The lineEN intersectsAF andMF atP andR. The lineMF intersectsCEatQ. Prove that if the quadrilateralsAMRP andCN RQhave inscribed circles, thenABCD has an inscribed circle (Fig.2.34).
Problem 2.64 The pointsA1, A2, C1andC2are chosen in the interior of the sides CD, BC, ABandADof the convex quadrilateralABCD. Denote byMthe point of intersection of the linesAA2 and CC1 and by N the point of intersection of the linesAA1andCC2. Prove that if one can inscribe circles in three of the four quadrilateralsABCD, A2BC1M, AMCN andA1N C2D, then a circle can be also inscribed in the fourth one.
Problem 2.65 A line cuts a quadrilateral with an inscribed circle into two polygons with equal areas and equal perimeters. Prove that the line passes through the center of the inscribed circle.
Problem 2.66 In the convex quadrilateralABCDwe have∠B=∠C=120◦, and AB2+BC2+CD2=AD2.
Prove thatABCDhas an inscribed circle.
Problem 2.67 LetABCD be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60◦. Prove that
1
3 AB3−AD3 ≤ BC3−CD3 ≤3 AB3−AD3 . When does equality hold?
zn=rn(cosnθ+isinnθ ) (known as de Moivre’s formula).
Now, ifz=cosθ+isinθ(that is, its absolute value equals 1) then 1
z= 1
cosθ+isinθ = cosθ−isinθ
cos2θ+sin2θ =cosθ−isinθ.
We obtain the useful formulas cosθ=1
2
z+1 z
, sinθ= 1 2i
z−1
z
. Moreover, using de Moivre’s formula, we have
cosnθ=1 2
zn+ 1
zn
, sinnθ= 1 2i
zn− 1
zn
.
These formulas may be very useful in solving a lot of Dr. Trig’s problems. We might even forget some of his formulas. For instance, we have
cos 2θ=1 2
z2+ 1
z2
=1 2
z+1 z
2
−1=2 1
2
z+1 z
2
−1=2 cos2θ−1.
Also 1
2
z+1 z
3
=1 8
z3+3z+3 z+ 1
z3
=1 8
z3+ 1 z3
+3
8
z+1 z
. We deduce
cos3θ=1
4cos 3θ+3 4cosθ, a formula sometimes written as
cos 3θ=4 cos3θ−3 cosθ.
Now, let us put some of this at work.
2.8 Dr. Trig Learns Complex Numbers 67