The harmonic quadrilateral and its properties Nikolaos Rapanos email: [email protected] HMS-Preparation Notes for I.M.O. 2009 July 7, 2009

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The basics

In the following we assume that the reader is familiar with properties of symmedian. Lemma 1.1. Given a cyclic quadrilateral ABCD and denoting by K the intersection point of its diagonals, the following relationship holds: AD · AB AK = BC · CD KC

Figure 1: Lemma for cyclic quadrilaterals.

Proof. Since ABCD is cyclic, ∠C = 180 − ∠A and therefore

(ABD) AB · BD = , but (BCD) BC · CD

h1 AK (ABD) = = and this finishes the proof. (BCD) h2 KC Lemma 1.2. If AL is a symmedian of triangle ABC and if D is the intersection of AL with the circumference, then DL is the symmedian of triangle BCD. Furthermore BL and CL is the symmedian of triangles ABD and ACD respectively. 1

Figure 2: A useful property of the symmedians.   BL BD 2 AB BD Proof. It is sufficient to show that = ⇐⇒ = . But from the previCL CD AC CD  2 AC · CD CL AC CD AC =⇒ ous lemma we deduce that = = = and we are done. AB · BD LB AB BD AB Alternatively one could consider the distances of point D from the lines AD, AC and use the fact that the A-symmedian is the locus of the points which are such that the ratio of their distances from the sides AB and AC is equal to the ratio of the sides.     BD 2 DL BD 2 BD CD · = =⇒ = and thus BL is the Now observing that AB AC AB LA AB symmedian of triangle ABD and the proof is now complete. Theorem 1.1. (Butterfly Theorem) Let M be the midpoint of a chord P Q of a circle, through which two other chords AB and CD are drawn; AD and BC intersect chord P Q at X, Y respectively. Then M is the midpoint of XY .

Figure 3: Butterfly Theorem.

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Proof. Let X 0 be the reflection of Y with respect to M . Then since both D and B lie on the circle, we have R(DP, DC; DA, DQ) = R(BP, BC; BA, BQ) therefore R(P, M ; X, Q) = 1 = R(P, M, X 0 , Q) and hence R(P, Y ; M, Q). But R(P, Y ; M, Q) = R(P, M ; Q, X 0 ) R(P, M, X 0 , Q) = R(P, M ; X, Q) which means that X ≡ X 0 and this proves that M is the midpoint of the segment XY . Remark. In the proof above we used the notation R(A, B; C, D) to indicate the cross-ratio of the four points which is defined as follows (for colinear points): R(A, B; C, D) =

−→ AC − − → CB

:

− − → AD −−→ DB

Since the intersections of any line with a bundle of four radii lead to identical cross-ratios, it is meaningful to define the cross-ratio for pairs of concurrent points as well as for concyclic points. Further information will be provided in the lectures, but students are strongly advised to study [1].

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On the harmonic quadrilateral

General Remark about notation. If the point A belongs to the circle ω, then in the following we use the notation AA to denote the tangent to ω at the point A. Theorem 2.1. Let ABCD be a convex cyclic quadrilateral. Let S = AC ∩ BD, E = AA ∩ CC, F = BB ∩ DD, AB = a, BC = b, CD = c and DA = d. The following properties are equivalent: 1. ac = bd. 2. E ∈ BD. 3. F ∈ AC. 4. The point S is the foot of a symmedians in each of the triangles ABD, ABC, BCD and ACD. If any of the above properties holds, then all of them hold and the quadrilateral ABCD is called harmonic quadrilateral. We have chosen the name harmonic quadrilateral because the first property can be written in the form

BA BC

=

DA DC

which can be thought as a

genralization of the harmonic conjugate points relationship for points belonging to a circle. Formally, this can be described using the concept of cross-ratio. Remark. Let M = AB ∩ CD and N = BC ∩ AD; then from Pascal’s Theorem one can deduce that M N coincides with the line EF .

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Figure 4: The Harmonic Quadrilateral. Proof. First of all we shall prove that 1 and 2 are equivalent. Assuming that F ∈ AC we   SD AD 2 have that AE is the A-symmedian of the triangle ADB and thus = . But SB AB SD CD · AD applying Lemma 1.1 we get = and equating the right handsides we deduce SB AB · BC that ac = bd. Since we used equivalences throughout the proof we conclude that 1 and 2 are indeed equivalent. Now propositions 2 and 3 are equivalent as a consequence of La Hire’s theorem. In particular, assuming that proposition 2 holds, we have that the polar of E (which is AC) passes through F and according to La Hire’s theorem the polar of F must also pass throuh E, hence E ∈ BD. For the validity of proposition 4, first notice that it is enough to show that AF is the A-symmedian of the triangle ABD since the rest are consequence of Lemma 1.2. Now assuming that proposition 2 is true, proposition 4 becomes a well-known theorem about symmedians. For completeness we give the following proof with referece to Figure 5. First of all consider the distances F T1 , F T2 , F T3 from the sides AD, DB, BA respectively. Observe that ∠F T2 T3 = ∠F BT3 = ∠ADB = ∠T1 F T2 therefore T2 T3 //T1 F . Similarly we can prove that T3 F//T1 T3 and therefore we deduce that T2 is the orthocenter of triangle AT1 T3 . Now one should observe that since ∠F T1 A = ∠F T3 A = 90 and hence AF passes through the circumcenter of the triangle AT1 T3 . This tells us that AF and AT2 are isogonal conjugate with respect to angle ∠A (well-known fact, but if you need further reading on isogonal conjugation you are advised to study [3]). In order to

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Figure 5: An interesting construction of a symmedian. finish the proof we only have to prove that AT2 is the A-median of triangle ABD but this is obvious since F T2 is the height of the isosceles triangle BF D. Theorem 2.2. Suppose that ABCD is a harmonic quadrilateral. Then (A, C; S, F ) = −1 and (B, D; S, E) = −1 with S, E and F as defined earlier. Proof. From the proof of proposition 4 of the previous theorem, one can deduce that BD is the polar of F and AC is the polar of E. The desired relationships can be proven using the well-known theorem about polars; that the pole together with the the three intersections of any line passing through the pole with its polar and the circle, define four harmonic conjugate points on the line. Theorem 2.3. Suppose that ABCD is a harmonic quadrilateral and O any point on the circumcircle of the quadrilateral. Then the bundle O(ABCD) is harmonic. Proof. Consider an inversion I(O, r2 ). If you are not familiar with the transformation of Inversion and its properties, you are advised to study [4] or [5]. A basic property of inversion is that any segment P Q under the inversion I is mapped to a segment P 0 Q0 r2 such that P 0 Q0 = · P Q. So considering the inversion I (and since the cenOP · OQ ter of inversion lies on the circle)the concyclic points A, B, C, D will be mapped to the colinear points A0 , B 0 , C 0 , D0 so that the segments that are created obey the above metric relationship. Since the quadrilateral ABCD is harmonic, we have 5

BA BC

=

DA DC .

So

Figure 6: Harmonic bundle derived from the harmonic quadrilateral B 0 A0 B0C 0

=

D 0 A0 D0 C 0

⇐⇒

now complete.

r2 ·BA OB·OA r2 ·BC OB·OC

=

r2 ·DA OD·OA r2 ·DC OD·OC

⇐⇒

BA BC

=

DA DC

which is true and the proof is

Alternatively, one could use the previous theorem according to which the bundle D(ABCD) is harmonic. Therefore if O is any point on the circumcircle, the bundle O(ABCD) will be equivalent with the bundle D(ABCD) and we are done.

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Applications

Problem 1. (BMO 2009) Let ABC be a triangle and consider an arbitrary parallel line to BC that cuts AB and AC at M and N respectively. Denote by P the intersection of BN and CM . If the circumcircles of the triangles BM P and CP N intersect at P and Q, prove that ∠BAQ = ∠CAP . Solution. (Nikolaos Rapanos) First of all it is easy to prove that AP is passing through the midpoints of M N and BC. You can prove this by using Thale’s Theorem or an alternative easy way to look at it, would be to consider the intersection of M N with BC, which of course is infinity, and then recall the well known theorem that the diagonals

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of a complete quadrilateral intersect into harmonic conjugate points, hence AP passes through the midpoint of BC (which is the h.c. of infinity). Note that Q is Miquel’s point for the complete quadrilateral AN P M . Having proved that, we now have to prove that AQ is the A-symmedian. Denote by D the intersection between the circumference of ABC and AQ. In order to prove that AQ is symmedian in the triangle ABC, we only have to prove that ACDB is a harmonic quadrilateral. Therefore it is enough to show that BD · AC = DC · AB. We will prove this relationship by considering two pairs of similar triangles. Observe that ∠QDB = ∠N CB and ∠BQD = 180 − ∠BQA = 180 − ∠AN B = ∠BN C, thus BQD ∼ BCN . Similarly we can prove that CQD ∼ CM B. Therefore BD BC BD CD

= =

QD NC MB NC

QD CD BC = M B . AB AC =⇒ q.e.d.

and

Now we divide the last two relationships by parts and we get

=



Problem 2. (Konstantinos Pappelis) Let ABCD be a quadrilateral inscribed in a circle centered at O and call M the intersection of AC and BD. Let P be a point on segment BC such that P M is perpendicular to M O. Let DP meet the circle at S. Let Q be a point of the circle such that DQ is perpendicular to OM . Denote by R the intersection of the bisectors of angles ∠ABS and ∠AQS and by L the intersection of the tangents through B and Q. Prove that A, R, S, L are colinear. Solution. (Nikolaos Rapanos) First of all prove on your own that it is sufficient to show that the quadrilateral ABSQ is harmonic. You will need the lemma of inscribed quadrilaterals according to which if ABCD is inscribed in a circle and T is the intersection point of its diagonals then

AT TC

=

AD·AB BC·CD .

Now redraw your configuration. You only have to

prove that the bundle D(A, B, S, Q) is harmonic. Now observe that if P M intersects AD at K, then according to butterfly theorem M P = M K. Therefore the bundle D(K, M, P, Q) is harmonic as desired. 

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Problems

Problem 3. (Nikolaos Rapanos) Let ABC be a triangle inscribed in a circle (O). The tangents to the circle at B and C intersect at P and the tangent at A intersects BC at L. We construct the parallelogram AN BM where M, N are points on BC and BP respectively. Denote by Q the intersection of AP and M N . Show that O, L and Q are collinear.

References [1] Milivoje Lukic. Projective Geometry, 2007. [2] http://www.mathlinks.ro/viewtopic.php?searchid=423494922t=43652 7

[3] Darij Grinberg. Isogonal Conjugation with respect to a triangle, version 3 September 2006. [4] Dusan Djukic. Inversion, 2007. [5] Zvezdelina Stankova-Frenkel. Inversion in the Plane-Part I, Berkeley Math Circle 1998-1999 [6] Cosmin Pohoata. Harmonic Division and its Applications.

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