Download APOLLONIUS GALLUS

A PO L L O N I U S G AL L U S
[Source: Francois Viete, Apollonius Gallus, Paris, 1600]
(The numbering of the lemmas has been modified.)
Problem 1 [PPP]
Given 3 arbitrary points, either they lies on a same circle, or there does not exists a straight line
passing through these 3 points.
Given 3 points A, B, C.
Join AB, BC, CA. If a triangle is formed, then they are not collinear.
By Elements(IV.5), there exists a circle passing through points A, B, C. ■
Lemma 1
In triangle ABC, cut AB at D such that AB  AD  AC 2 . Then the points D, B, C can form a circle
DBC which touches AC.
Assume the contrary that AC is not a tangent to the circle DBC.
Then AC cuts the circle at 2 points, let the other point be E.
Since AB  AD  AC  AE (Elements III.36),
which is impossible because AC  AE , so AB  AD  AC  AE  AC 2 .
Therefore, AC is not a secant but tangent to the circle DBC. ■
Problem 2 [PPL]
Given 2 points and a straight line, construct a circle passing through these 2 points and touching the
given straight line.
(Case 1)
Given 2 points A, B and a straight line
(CZ).
If AB is parallel to the straight line ,
Bisect AB at D, drop D to the straight line
at C. (that means ADC  90 )
Construct the circle ABC.
Since AB CZ , and ADC  90 , so DC  CZ .
Therefore, the circle ABC touches
(Case 2)
Given 2 points A, B and a straight line
If AB is not parallel to the straight line
Produce AB to
at C. ■
(EZ).
,
at E, construct a point C on
Construct the circle ABC.
By Lemma 1, the circle ABC touches
at C. ■
such that EB  EA  EC 2 .
Problem 3 [LLL]
Given 3 straight lines, construct a circle touching the 3 given straight lines. (It is necessary that no
tow of the 3 straight lines are parallel to each other.)
(Remark: the proof is the same as Elements IV.4)
Let B, D be 2 intersections.
Now bisect angle B and angle D. The 2 angle bisectors intersect at E.
Fall from E (3 perpendiculars) to the 3 given straight lines at A, F, C.
Construct the circle AFC.
Since EAB  EFB , so EA = EF.
Also, EFD  ECD , so EF = EC.
Therefore, EA = EF = EC and E is the center of the circle AFC, which touches the 3 given straight
lines at A, F, C. ■
Lemma 2
To draw a line passing through a given point such that it intersects 2 given lines at equal angles.
(case 1)
(case 2)
Given a point A and 2 straight lines BC, DE.
Case 1:
If BC is parallel to DE, then fall perpendicular to BC and DE, intersecting both at right angles.
Case 2:
If BC is not parallel to DE, then the 2 lines intersect (at F).
Bisect DFB .
Fall perpendicular from A to this angle bisector at G, intersecting BF and DF at H, I respectively.
Since HGF  IGF (ASA: HGF  IGF  90 , GF common side, HFG  IFG ),
therefore GHF  GIF . ■
Problem 4 [PLL]
Given 2 straight lines and a point, construct a circle passing through the given point and touching the
2 given straight lines.
Given a point A and 2 lines BC and DE.
By Lemma 2, through A draw a line intersecting BC, DE at H, I respectively with equal angles.
So, we have the point K, the intersection of angle bisector and the perpendicular from A, and the
point L, the reflection of A about the angle bisector (AK = KL).
Construct a point M on BC such that HL  HA  HM 2 .
Construct the circle ALM.
Obtain the centre N, fall perpendicular to DE at O.
(Prove the circle ALM touches BC and DE at M and O respectively, i.e. NM = NO)
Since NKH  NKI (SAS: NK = NK, NKI  NKH  90 , NH = NI),
Therefore NHK  NIK and NH = NI.
So, NHM  NIO ( NHM  KHM NHK  KIO NIK  NIO ).
Also, NMH  NOI  90 .
 NKH  NKI (ASA), it follows that MN = NO. ■
Problem 5 [CLL]
Given a circle and 2 straight lines, construct a circle touching the given circle and the 2 given straight
lines.
Given a circle with centre A and two straight line ZC, DB.
Translate ZC, DB parallel to XH, FG respectively by the radius of the given circle. (Viete’s word:
perp. to AZ, AD resp).
The dotted circle E shall passes through A and touches both lines XH and FG.
By Problem 4 [PLL], construct the circle passing A and touching XH, FG.
(The proof) Simply reverse the above steps. ■
Lemma 3
If two circles are mutually secant (intersect at two distinct points), one’s diameter must not pass
through the center of the other circle.
Given that the circles CEB and CED are mutually secant (not mutually tangent), they must intersect
at two points, C and E, say.
Let A be the centre of circle CEB.
Produce CA to meet the circle A at B and the other circle at D.
Claim that the line CABD does not pass through the circle of the circle CED.
Join EC, EB, ED.
So, in the semi-circle, CEB  90 .
Plus or minus the angle BEB , therefore CED  90 .
Hence CABD is not the diameter of the circle CED, and therefore does not pass through the centre. ■
Lemma 4
If two circles intersect (not mutually tangent), from an intersecting point, draw a line cutting the two
circles, the two arcs of the cutting segments are not similar (with respect to their circles).
Given two circles CEB and CED are mutually secant, intersect at C, E.
The line CBD cuts the circle CEB at C, B and cuts the circle CED at C, D.
Claim: arcs CB and CD are not similar.
Case 1:
If CB passes through the center of CEB (A),
Then CB does not pass through the center of the circle CED (by Lemma 3).
Now CB is a diameter of circle CED, but not a diameter of circle CED; therefore CE, CD are not
similar arcs.
Case 2:
If CB does not pass through the center of CEB (A),
produce CA to meet the circle CEB ar G and circle CED at F. Join BG, DF.
In semi-circle CBE, CBG  90 .
On the other hand, CDF  90 . (since CG is a diameter of circle CBE, so CF is not a diameter of
circle CED.)
Therefore, BG is not parallel to DF.
Let H be the center of the circle CED, and CHI be a diameter.
Produce DI and CG to meet at K.
Therefore BG DK . ( CBG  CDI  90 )
For CB : CD  CG : CK  CG : CI (CI  CK )
So, CB and CD are not similar. ■
Lemma 5
If two triangles have parallel bases, and share a same top vertex, then the 2 circumscribed circles are
mutually tangent to each other.
Let triangles ABD and AEF share the same vertex A and BD EF .
Construct the 2 circumscribed circles ABD and AEF.
The 2 circles will either be mutually tangent or mutually secant to each other.
Since ABD AEF , so AB : AE is the ratio of the radii ratio of the two circles, hence AB and AE
are similar (arcs).
By Lemma 4, the two circles are mutually tangent to each other. ■
Problem 6 [CLP]
Given a point, a straight line and a circle, construct a circle passing the given point and touching the
given line and circle.
Given a point A, a straight line (BC) and a circle (DEF).
Let G be the center of the circle and fall perpendicular to the given line such that DGFC is
perpendicular to BC  DC , and DF is a diameter.
Join DA.
Construct the point H on DA such that DA DH  DC  DF .
Therefore, A, H, C, F are concyclic.
Construct the circle passing through A, H and touching the given line by Problem 2 [PPL] at B.
Join DB and cuts the given circle at E. Join FE
Now, DEF  BCF  90 .
Thus, since the sum of opposite angles in the quadrilateral BEFC is DEF  BCF  180 ,
B, E, F, C are concyclic,
so DB  DE  DF  DC  DA DH , therefore, A, H, E, B are concyclic.
(Prove that the circle DEF touches the circle AHEB).
First, construct the diameter BI.
Since this circle touches the given line at B, CBI  90 , and IB DC .
Also, IEB  90  DEF and DEB is a straight line, so IEF is also straight.
Note that the triangles DEF and BEI are similar, sharing the same top vertex E with parallel bases,
So their circumscribed circles touches each other at E, (but not secant).
In conclusion, the circle BAE touches the given line at B and touches the given circle at E. ■
Problem 7 [CCL]
Given two circles and a straight line, construct a circle touching the 3 given objects.
Given two circles with center A and B, and a straight line (CZ).
Perform parallel translation to the given line in the direction perpendicular to BZ (away), by a unit
equal to the radius ( R1  AL ) of the first circle. A new line HX (dotted) is formed.
Construct a circle concentric to B, with radius equals the difference of the radii of the first and
second given circles ( BG  r  R1  R2 )
Construct the (dotted) circle (AGH), touching the new circle at G, touching the new line at H and
passing through A. The center is then called E.
Reduce the radius of dotted circle by R1  AL , we form a circle LMC with center E, which touches
the 2 given circles A and B (both internally), and the given line at C.
Alternative cases:
Case I
To touches both externally,
When R1  R2 ,
Take BG  r  R1  R2 .
CZ translate the way back perp. to BX.
Case II
To touches both internally,
When R1  R2 ,
Take BG  r  R1  R2 .
CZ translate away perp. to BX.
Case III
To touches one internally and the other
externally,
Take BG  r  R1  R2 .
CZ translate the way back perp. to BX.
As mentioned above
Problem 8 [PPC]
Given two points and a circle, construct a circle passing the two given points and touching the given
circle.
(Remark: For the first diagram, H lies on DB and HF , BK are tangents to A.)
Given two points B, D and a circle A.
Join BD. Construct a point H on BD such that BD  BH  AB 2  AF 2 .
Construct the tangent from H to A, touching at F.
Produce BF to meet the circle at G.
Join DG to intersect the circle at E. Join EF.
Construct the circle GBD.
(Prove that the circle GBD touches the circle A.)
Claim that GDB
GEF , i.e. to prove GDB  GEF .
Now, BD  BH  AB 2  AF 2  BK 2  BG  BF , so D, G, H, F are concyclic.
So, GDB  HFB .
Also, EGF  EFH since HF is a tangent to circle A.
Therefore, GEF  180  EGF  GFE  180  EFH  GFE  HFB  GDB .
So, GDB GEF with parallel bases and having the same vertex G.
The two circumscribed circles DBG and GEF are mutually tangent at G. ■
Lemma 6
(Homothetic center)
Given two circles, find the point on the line joining the centers, from which when a straight line is
drawn cutting the two circles, the segments will be similar.
Given two circles, the first one ABC with center K, the second one EFG with center L.
Join KL.
Construct the point M on KL such that KM : LM  AK : EL .
Then M is the homothetic center.
Let the line MGFCB cuts the circle ABC at B, C; and cuts the circle EFG at F, G;
B, F are points on the same side; C, G are points nearer to M.
Join BK, CK, FL, LG. And so triangles BKC, FLC are formed.
Therefore by construction,
KM : LM  KB : FL , BK FL , KBC  LFG .
Similarly,
KM : LM  KC : LG , KC LG , KCB  LGF .
So, the third angles must be equal, i.e. BKC  FLG .
Hence segments BC, FG are similar. ■
(M can be inside or outside KL.)
Lemma 7
Given two circles ABCD, EFGH, the line joining the centers KL cuts the circles at A,D, E, H, where
AD is a diameter of the first circle and EH is a diameter of the second, M is constructed by Lemma 6.
MGFCB cuts the first circle at B, C; cuts the second circle at F, G (hence similar arcs). In the first
circle, A, B are farther points and C, D are nearer to M; in the second circle, F, E are farther and G, H
are nearer.
Then
MG  MB  MH  MA ……(1)
Also
ME  MD  MF  MC ……(2)
(1)
Fall perpendiculars from K to BC (to R); from L to FG (to S). Join KC, LG.
BC and FG are similar.
By the half, RC and SG are similar.
Therefore, KRC LSG and KC LG .
Even, CKD  GLH and finally the subtended chords are parallel CD GH .
Therefore,
MD : MC  MH : MG .
MD : MC  MB : MA . (Remark: MDC
MG  MB  MH  MA .
(2)
In circle EFGH,
But
Therefore
MH : MG  MF : ME ,
MH : MG  MD : MC .
MF : ME  MD : MC , hence ME  MD  MF  MC . ■
So,
But
MBA by AAA)
Problem 9 [CCP]
Given two circles and a point, construct a circle passing through the given point and touching the two
given circles.
(Remark: H does not necessarily lie on the desired circle.)
Given two circles ABCD (center K), EFGH (center L) and a point I.
Join KL and cuts circles such that AD, EH are diameters of the first and second circles respectively.
Also obtain the homothetic center M.
Join MI and construct the point N on MI such that MI  MN  MH  MA .
Construct the circle passing through I, N and touching the circle ABCD. (by Problem 8 PPC)
Let the point of contact be B and join BM, cutting the circles ABCD and EFGH at B, C and F, G
respectively.
Since
(by Lemma 7)
MG  MB  MH  MA
By construction,
(N, I, B, G concyclic)
MG  MB  MN  MI
But G also lies on the circle EFGH, so the two circles EFGH and IBGN are either mutually secant or
mutually tangent to each other.
Since the two circles BNI and BCD are mutually tangent to each other. (Remark: Here B becomes
the homothetic center.)
For BG and BC are similar, and FG and BC are similar; thus FG and BG are similar.
Therefore, the two circles EFGH and IBGN are mutually tangent to each other.
In conclusion, the circle IBG is the desired one which passes through the given point I and touching
the two given circles ABD, EGH at B, G respectively. ■
Problem 10 [CCC]
Given three circles, construct a circle touching the three given circles.
Given a first circle A , second B, third D (remark: with radii R1 , R2 , R3 respectively)
Construct a circle centered at D with radius DF  R1  R3 .
Construct a circle centered at B with radius BG  R1  R2 .
Lastly, through A construct a circle AGF which touches the two newly constructed circles at G, F;
and that this circle has a center E.
Increase the radius by R1 , the circle HLM then satisfies the requirement.
Other cases:
Case (I)
To touches 3 given circles internally.
Let R1  R2  R3 .
Choose
DF  R1  R3 , BG  R1  R2 .
Case (II)
To touches 3 given circles externally.
Let R1  R2  R3 .
Choose
DF  R1  R3 , BG  R1  R2 .
Case (III)
To touches A internally; and B, D
externally.
Choose
DF  R1  R3 , BG  R1  R2 .
As shown above
Case (IV)
To touches A externally; and B, D
internally.
Choose
DF  R1  R3 , BG  R1  R2 .