holds if and only if the sum of the carpets’ areas equals the area ofABC. But this is very simple to prove if we notice that (Fig.5.47)
ABB
= [ABM], BCC
= [BCM]
and
CAA
= [CAM]. We obviously have
[ABM] + [BCM] + [CAM] = [ABC].
5.7 Quadrilaterals with an Inscribed Circle
Problem 2.60 Prove that if in the quadrilateralABCD is inscribed a circle with centerO, then the sum of the angles∠AOBand∠CODequals 180◦(Fig.5.48).
Solution We know that the quadrilateral’s angles bisectors intersect atO. Thus
∠AOB=180◦−∠ABO−∠BAO=180◦−∠A+∠B
2 .
Similarly,
∠COD=180◦−∠C+∠D
2 .
Adding these equalities, we obtain
∠AOB+∠COD=360◦−∠A+∠B+∠C+∠D
2 =360◦−360◦
2 =180◦.
Fig. 5.48
Fig. 5.49
Problem 2.61 LetABCDbe a quadrilateral with an inscribed circle. Prove that the circles inscribed in trianglesABCandADCare tangent to each other.
Solution Suppose that the circles inscribed in trianglesABC andADCtouchAC at the pointsXandY, respectively. We have to prove thatX=Y. With the notation in Fig.5.49, we haveAX=AK (the tangents to a circle drawn from a point are equal),BK=BL,CL=CX,CY=CM,DM=DN,AN=AY.
BecauseABCDhas an inscribed circle,AB+CD=AD+BC, so AK+BK+CM+MD=AN+DN+BL+CL.
5.7 Quadrilaterals with an Inscribed Circle 187 Fig. 5.50
Using the previous equalities, we obtain
AX+CY=AY+CX.
Adding the obvious equality
AX+XC=AY+Y C yields 2AX=2AY, henceX=Y.
Observation It can be proven that the pointsK, L, M, Nare the vertices of a cyclic quadrilateral (Fig.5.50).
Indeed, if we draw the circle inscribed inABCDand denote byK, L, Mand N the tangency points with the quadrilateral’s sides, it is not difficult to see that the sides ofKLMN andKLMNare parallel, hence their corresponding angles are equal. ButKLMN is cyclic, thus its opposite angles add up to 180◦. The conclusion follows (Fig.5.51).
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:
(i) a circle is inscribed inABCD; (ii) BE+BF=DE+DF; (iii) AE−AF =CE−CF.
Solution Suppose a circle is inscribed in the quadrilateralABCDand touches its sides at the pointsK, L, M, N (Fig.5.52).
Observe that
BE+BF=EK−BK+BL+LF=EM+N F
=EM+N D+DF=EM+MD+DF =DE+DF.
We used again the fact that the tangents to a circle from a point are equal, so BK=BL,EK=EM,LF=N F andN D=MD.
In a similar way we have
AE−AF =AK+EK−AN−N F=EK−N F
=EM−LF=CE+CM−CL−CF=CE−CF.
Conversely, if, for instance,BE+BF =DE+DF, draw the circle tangent toAB, BC, and AF. If this circle is not tangent toCD as well, draw fromE a tangent to the circle which intersectsAF atD. Then BE+BF=DE+DF, and we deduceD=D, a contradiction. We conclude thatABCDhas an inscribed circle.
Problem 2.63 Let ABCD be a convex quadrilateral. Suppose that the lines AB andCD intersect atEand the linesADandBCintersect atF. LetM andN 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.
5.7 Quadrilaterals with an Inscribed Circle 189
Fig. 5.52
Fig. 5.53
Solution Suppose the points are located as in Fig.5.53, the other cases being simi- lar. BecauseAMRP has an inscribed circle, it follows from the preceding problem thatAE+AF =RE+RF. Analogously, sinceCN RQhas an inscribed circle, we haveRE+RF=CE+CF. We obtainAE+AF =CE+CF and this implies thatABCDhas an inscribed circle.
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.
Solution Letα=AB−BC−AM+CM. From the previous problem it follows that a circle can be inscribed in the quadrilateralA2BC1Mif and only ifα=0.
Analogously, if we set
β=CD−AD−CN+AN,
thenβ=0 if and only ifA1DC2N has an inscribed circle. Setting γ=AM−CM+CN−AN,
δ=BC−AB+AD−CD,
from the theorem of Pithot it follows thatγ=0 andδ=0 are necessary and suf- ficient conditions for the existence of an inscribed circle inAMCN andABCD, respectively.
Now, simply observe thatα+β+γ+δ=0, so that if three of the four numbers are zero, then so is also the fourth one. It follows that if one can inscribe circles in three of the four quadrilaterals, a circle can also be inscribed in the fourth one (Fig.5.54).
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.
Solution The line cuts the quadrilateral either into two quadrilaterals, or into a tri- angle and a pentagon. The reasoning is basically the same in both cases, so we as- sume that the line intersect the sidesABandCDat the pointsXandY (Fig.5.55).
LetO be the center of the inscribed circle. BecauseAXY D andBXY C have the same perimeter, it follows that
AX+AD+DY=BX+BC+CY.
Multiplying this equality with12R(Rbeing the radius of the inscribed circle) yields [OAX] + [OAD] + [ODY] = [OBX] + [OBC] + [OCY],
5.7 Quadrilaterals with an Inscribed Circle 191
Fig. 5.55
that is,
[OXADY] = [OXBCY]. Because
[OXADY] + [OXBCY] = [ABCD], it follows that
[OXADY] = [OXBCY] =1
2[ABCD].
Now, suppose by way of contradiction that the lineXY does not pass throughO.
SupposeOlies in the interior ofAXY D(the case in whichO is in the interior of BXY Cis similar). We have
[BXY C] =1
2[ABCD] = [BXOY C],
but[BXOY C] = [BXY C] + [OXY], hence[OXY] =0, which is a contradiction.
Problem 2.66 In the convex quadrilateralABCDwe have∠B=∠C=120◦, and AB2+BC2+CD2=AD2.
Prove thatABCDhas an inscribed circle.
Solution SupposeABandCDintersect atE. Since∠B=∠C=120◦, the triangle BCE is equilateral. Denote by x its side length. Applying the law of cosines in
(AB+CD−x) =AB +CD +x −2(ABãx+CDãx−ABãCD)
=AB2+CD2+x2
=AD2. It follows that
AB+CD=AD+x=AD+BC, thereforeABCDhas 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
3AB3−AD3≤BC3−CD3≤3AB3−AD3. When does equality hold?
Solution By symmetry, it suffices to prove the first inequality.
SinceABCDhas an inscribed circle, we haveAB+CD=AD+BC, or, equiv- alently,AB−AD=BC−CD. Therefore, the inequality we want to prove is equiv- alent to
1 3
AB2+ABãAD+AD2
≤BC2+BCãCD+CD2.
From the hypothesis we have 60◦≤∠A,∠C ≤120◦, therefore 12 ≥cosA, cosC≥ −12. Applying the law of cosines in triangleABDyields
BD2=AB2−2ABãADcosA+AD2≥AB2−ABãAD+AD2. But
AB2−ABãAD+AD2≥1 3
AB2+ABãAD+AD2
, (5.3)
the latter being equivalent to
(AB−AD)2≥0.
Applying again the law of cosines in triangleBCDyields
BD2=BC2−2BCãCDcosC+CD2≤BC2+BCãCD+CD2. (5.4)