Forum Geometricorum V olume 10(2010)141–148.
ISSN 1534-1178
Constructions with Inscribed Ellipses in a Triangle
Nikolaos Dergiades
Abstract . We give a simple construction of the axes and foci of an inscribed
ellipse with prescribed center, and as an application, a simple solution of the
problem of construction of a triangle with prescribed circumcevian triangle of
the centroid.
1. Construction of the axes and foci of an inscribed ellipse
Given a triangle ABC , we give a simple construction of the axes and foci of an inscribed ellipse with given center or perspector. If a conic touches the sides BC , CA , AB at X , Y , Z respectively, then the lines AX , BY , CZ are concurrent at a point P called the perspector of the conic. The center M of the conic is the com-plement of the isotomic conjugate of P . In homogeneous barycentric coordinates, if P =(p :q :r ) , then 111111+:+:+=(p (q +r ) :q (r +p ) :r (p +q )) .
M =q r r p p q
B Q B Figure 1
Consider triangle ABC and its inscribed ellipse with center M =(u :v :w ) as the orthogonal projections of a triangle AB C and its incircle. We may take A =A and assume BB =m , CC =n . If the incircle of AB C touches B C , C A and AB at X , Y , Z respectively, then X , Y , Z are the orthogonal projections of X , Y , Z . It follows that the perspector P is the projection of Publication Date:November 15, 2010. Communicating Editor:Paul Yiu.
142N. Dergiades the Gergonne point G e of AB C . Suppose triangle AB C has sides B C =a , C A =b and AB =c , then
111b +c −a :c +a −b :a +b −c =::. p q r
It follows that
111111a :b :c =+:+:+=u :v :w. q r r p p q
From this we draw a remarkable conclusion.
Proposition 1. If a triangle and an inscribed ellipse are the orthogonal projections of a triangle and its incircle, the sidelengths of the latter triangle are proportional to the barycentric coordinates of the center of the ellipse.
Since a , b , c satisfy the triangle inequality, the center M =(u :v :w ) of the ellipse is an interior point of the medial triangle of BAC .
We determine the relative positions of ABC and AB C leading to a simple construction of the axes and foci of the inscribed ellipse of given center M . We firstconstruct two triangles A +BC and A −BC with sidelengths proportional to the barycentric coordinates of M .
Construction 2. Let M =(u :v :w ) be a point in the interior of the medial triangle of ABC , with cevian triangle A 0B 0C 0. Construct
(i)the parallels of BB 0and CC 0through A to intersect BC at D and E respec-tively,
(ii)the circles B (D ) and C (E ) to intersect at two points A +and A −symmetric in the line BC . Label A +the one on the opposite side of BC as A .
Each of the triangles A +BC and A −BC have sidelengths u :v :w .
D E Figure 2
Constructions with inscribed ellipses in a triangle 143Proof. Note that
BC 0u BC BC ==, =CA −CE C 0A v BC B 0C u BC ===. A −B DB AB 0w
From these, BC :CA −:A −B =u :v :w ; similarly for BC :CA +:A +B . Lemma 3. The lengths of AA +and AA −are given by √√Q +16ΔΔ Q −16ΔΔ √√AA +=and AA −=, 2u 2u
where
Q =(b 2+c 2−a 2) u 2+(c 2+a 2−b 2) v 2+(a 2+b 2−c 2) w 2,
and Δ is the area of the triangle with sidelengths u , v , w .
Proof. Applying the law of cosines to triangle ABA −, we have
22AA 2−=AB +BA −−2AB ·BA −cos(ABC −A −BC ) aw 2aw 2(cosB cos A −BC +sin B sin A −BC ) −2c ·=c +u u c 2u 2+a 2w 2−2ca cos B ·wu cos A −BC −2ca sin B ·wu sin A −BC =u 2
2c 2u 2+2a 2w 2−(c 2+a 2−b 2)(w 2+u 2−v 2) −16ΔΔ =2u 2
Q −16ΔΔ =. 2u 2
The case of AA +is similar. (1)
Let Q be the intersection of the lines BC and B C . The line AQ is the inter-section of the planes ABC and AB C . The diameters of the incircle of AB C parallel and orthogonal to AQ project onto the major and minor axes of the in-scribed ellipse.
Proposition 4. The line AQ is the internal bisector of angle A +AA −.
ab Proof. The sidelengths of triangle A −BC are BC =a , CA −=av u =a and
ac =A −B =aw u a . Set up a Cartesian coordinate system so that A =(x 1, y 1) , B =(−a, 0) , C =(0, 0) , A +=(x 0, −y 0) and A −=(x 0, y 0) , where
a 2+b 2−c 2·a, x 0=−CA −cos A −CB =−2a 2
a 2+b 2−c 2. x 1=−b cos C =−2a Since the lines AA +and AA −have equations
(y 1+y 0) x −(x 1−x 0) y −(x 1y 0+x 0y 1) =0,
(y 1−y 0) x −(x 1−x 0) y +(x 1y 0−x 0y 1) =0, (2)(3)
144N. Dergiades a bisector of angle A +AA −meets BC at a point with coordinates (x, 0) satisfying
((y 1+y 0) x −(x 1y 0+x 0y 1)) 2((y 1−y 0) x +(x 1y 0−x 0y 1)) 2=. (y 1−y 0) 2+(x 1−x 0) 2(y 1+y 0) 2+(x 1−x 0) 2
Simplifying this into
2222222(x 1−x 0) x 2+(x 20+y 0−x 1−y 1) x +x 0(x 1+y 1) −x 1(x 0+y 0) =0,
and making use of (2)and (3),we obtain 2 2 2a +b 2−c 2a 2+b 2−c 2b b −−x 2+a x 222a 2a a 2a 2 2+b 2−c 22+b 2−c 2a a 2−b ·=0. −b 2·2a 22a 2
Now, since
a 2=(m −n ) 2+a 2,
we reorganize (4)into the form b 2=n 2+b 2, c 2=m 2+c 2, (4)(5)
((m −n ) x −na )((m (a 2+b 2−c 2)+n (c 2+a 2−b 2)) x −(n (b 2+c 2−a 2) −2mb 2) a ) =0. Note that the two roots
na X 1=m −n
satisfy
m 2b 2−mn (b 2+c 2−a 2) +n 2c 2>0, (x 0−x 1)(X 1−X 2) =a 2+(m −n ) 2
since the discriminant of m 2b 2−mn (b 2+c 2−a 2)+n 2c 2(asa quadratic form in m , n ) is (b 2+c 2−a 2) 2−4b 2c 2=−4b 2c 2sin 2A X 2and X 1is the root that corresponds to the internal bisector. Since X 1n X 1+a =m , this intercept is the intersection Q of the lines BC and B C .
Theorem 5. The inscribed ellipse of triangle ABC with center M =(u :v :w ) (insidethe medial triangle) has semiaxes given by
a max =u (AA ++AA −)
2(u +v +w ) and a min =u (AA +−AA −) . 2(u +v +w ) and (n (b 2+c 2−a 2) −2mb 2) a X 2=m (a 2+b 2−c 2) +n (c 2+a 2−b 2)
Proof. Making use of (5),we have
2mn =m 2+n 2−(m −n ) 2=(b 2+c 2−a 2) −(b 2+c 2−a 2) .
It follows that
4(b 2−b 2)(c 2−c 2) =((b 2+c 2−a 2) −(b 2+c 2−a 2)) 2,
Constructions with inscribed ellipses in a triangle 145and
4b 2c 2−(b 2+c 2−a 2) 2+4b 2c 2−(b 2+c 2−a 2) 2
=4b 2c 2+4c 2b 2−2(b 2+c 2−a 2)(b 2+c 2−a 2)
=2((b 2+c 2−a 2) a 2+(c 2+a 2−b 2) b 2+(a 2+b 2−c 2) c 2) . (6)Note that 4b 2c 2−(b 2+c 2−a 2) 2=4b 2c 2(1−cos 2A ) =4b 2c 2sin 2A =16Δ2and similarly, 4b 2c 2−(b 2+c 2−a 2) 2=16Δ(AB C ) 2. Let ρand ρ be the inradii of triangle AB C and the one with sidelengths u , v , w . These two triangles ρ(u +v +w ) ρ, we have have ratio of similarity ρ =2Δ
Δ(AB C ) =Δ ρ(u +v +w )
2Δ 2ρ2(u +v +w ) 2=. 4Δ
With this, (6)can be rewritten as
2(u +v +w ) 4ρ4−(u +v +w ) 2Q ρ2+32Δ2Δ 2=0.
This has roots ±ρ1and ±ρ2, where
√√Q +16ΔΔ +Q −16ΔΔ √, ρ1=22(u +v +w )
Note that ρ1ρ2=4ΔΔ , (u +v +w ) 2√ρ2=√Q +16ΔΔ −Q −16ΔΔ √. 22(u +v +w ) and
4ΔΔ Δρ1ρ2a min ===. a max Δ(AB C ) ρ2(u +v +w ) 2ρ2
Since ρ1>ρ2, it follows that a max =ρ=ρ1and a min =ρ2.
Now we construct the axes and foci of an inscribed ellipse.
Construction 6. Let M be a point in the interior of the medial triangle of ABC , and A +BC , A −BC triangles with sidelengths proportional to the barycentric co-ordinates of M (seeConstruction 2) . Construct
(1)the internal bisector of angle A +AA −to intersect the line BC at Q ,
(2)the parallel of AQ through M . This is the major axis of the ellipse .
Further construct
(3)the orthogonal projection S of Q on AA +,
(4)the parallel through M to AA +to intersect A 1S at T , (5)the circle M (T ) . This is the auxiliary circle of the ellipse .
Finally, construct
(6)the perpendiculars to the sides at the intersections with the circle M (T ) to intersect the major axis at F and F . These are the foci of the ellipse. See Figure 3.
146N. Dergiades
Figure 3.
2. The circumcevian triangle of the centroid
For the Steiner inellipse with center G =(1:1:1) , the triangles A +BC and A −BC are equilateral triangles, and √√222222a +b +c +43Δa +b +c −43Δand AA −=. AA +=22
By Theorem 5,
AA ++AA −AA +−AA −and a min =. a max =66
Theorem 7. Let G be the centroid of ABC , with circumcevian triangle A 1B 1C 1. The pedal circle of G relative to A 1B 1C 1has center the centroid G 1of A 1B 1C 1. Hence, G is a focus of the Steiner inellipse of triangle A 1B 1C 1.
Proof. Let a , b , c and a 1, b 1, c 1be the sidelengths of triangles ABC and A 1B 1C 1respectively. Let x =AG , y =BG and z =CG . By Apollonius’theorem,
2b 2+2c 2−a 2, x =922c 2+2a 2−b 2y =, 922a 2+2b 2−c 2z =. 92
Constructions with inscribed ellipses in a triangle
147
C 1
Figure 4
If P is the power of G relative to circumcircle of ABC , then
GA 1=P , x GB 1=P , y GC 1=P . z
From the similarity of triangles GBC and GC 1B 1, we have
GC 1CG ·GC 1P a 1===. a BG BG ·GC yz
From this we obtain a 1, and similarly b 1and c 1:
a 1=a P , yz b 1=b P , zx c 1=c P . xy (7)
The homogeneous barycentric coordinates of G relative to A 1B 1C 1are
ΔGB 1C 1:ΔGC 1A 1:ΔGA 1B 1
ΔGB 1C 1ΔGC 1A 1ΔGA 1B 1::=ΔGBC ΔGCA ΔGAB GB 1·GC 1GC 1·GA 1GA 1·GB 1::=GB ·GC GC ·GA GA ·GB (BG ·GB 1)(CG ·GC 1) (CG ·GC 1)(AG ·GA 1) (AG ·GA 1)(BG ·GB 1) ::=BG 2·CG 2CG 2·AG 2AG 2·BG 2P 2P 2P 2=22:22:22y z z x x y
=x 2:y 2:z 2.
The isogonal conjugate of G (relativeto triangle A 1B 1C 1) is the point
a 2b 2c 211=a 2:b 2:c 2. G =2:2:12x y z ∗
148N. Dergiades We findthe midpoint of GG ∗by working with absolute barycentric coordinates. 222Since x 2+y 2+z 2=13(a +b +c ) , and
2(a 2+b 2+c 2) , 3x +a =3y +b =3z +c =3we have for the firstcomponent, 3x 2+a 2x 21a 21==. +2x 2+y 2+z 2a 2+b 2+c 22(a 2+b 2+c 2) 3222222
Similarly, the other two components are also equal to 13. It follows that the midpoint ∗of GG is the centroid G 1of A 1B 1C 1. As such, it is the center of the pedal circle of the points G and G ∗, which are the foci of an inconic that has center and perspector
G 1. This inconic is the Steiner inellipse of triangle A 1B 1C 1.
3. A construction problem
Theorem 7gives an elementary solution to a challenging construction problem in Altshiller-Court [1,p.292, Exercise 11].The interest on this problem was reju-venated by a recent message on the Hyacinthos group [2].
Problem 8. Construct a triangle given, in position, the traces of its medians on the circumcircle.
Solution. Let a given triangle A 1B 1C 1be the circumcevian triangle of the (un-known) centroid G of the required triangle ABC . We construct the equilateral triangles A +B 1C 1and A −B 1C 1on the segment B 1C 1. Let G 1be the centroid of A 1B 1C 1and r =16(A 1A ++A 1A −) . The circle G 1(r ) is the auxiliary circle of the Steiner inellipse, hence the pedal circle of G (oneof the foci) relative to A 1B 1C 1. From the intersections of this circle with the sides of A 1B 1C 1and the parallel from G 1to the internal bisector of angle A +A 1A −we determine the point G (twosolutions). The second intersections of the circle with the lines A 1G , B 1G , C 1G give the points A , B , C .
References
[1]N. Altshiller-Court, College Geometry , second edition, Barnes and Noble, 1952; Dover reprint,
2007.
[2]Hyacinthos message 19264, September 14, 2010.
Nikolaos Dergiades:I. Zanna 27, Thessaloniki 54643, Greece
E-mail address :ndergiades@yahoo.gr