Construction of Malfatti Squares

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1 Forum eometricorum Volume 8 (008) FORUM EOM ISSN onstruction of Malfatti Squares Floor van Lamoen and Paul Yiu bstract. We give a very simple construction of the Malfatti squares of a triangle, and study the condition when all three Malfatti squares are inside the given triangle. We also give an extension to the case of rectangles.. Introduction The Malfatti squares of a triangle are the three squares each with two adjacent vertices on two sides of the triangle and the two remaining adjacent vertices from those of a triangle in its interior. We borrow this terminology from [3] (see also [, p.48]) where the lengths of the sides of the Malfatti squares are stated. In Figure, the Malfatti squares of triangle are Z a Y a, X b Z b and Y c X c. We shall call the Malfatti triangle of, and present a simple construction of from a few common triangle centers of. Specifically, we shall make use of the isogonal conjugate of the Vecten point of. This is a point on the rocard axis, the line joining the circumcenter O and the symmedian point K. Y a Z a Y c Z b X b Figure. X c Theorem. Let P be the isogonal conjugate of the Vecten point of triangle. The vertices of the Malfatti triangle are the intersections of the lines joining the centroid to the pedals of the symmedian point K and the corresponding vertices to the pedals of P on the opposite sides of. See Figure. Publication Date: March 0, 008. ommunicating Editor: Jean-Pierre Ehrmann. The Vecten point of a triangle is the perspector of the (triangle whose vertices are) the centers of the squares erected externally on the sides. It appears as X 485 in [6]. See Figure 6. Its isogonal conjugate appears as X 37, and is also called the Kenmotu point. It is associated with the construction of a triad of congruent squares in a triangle. In [5, p.68] the expression for the edge length of the squares should be reduced by a factor. correct expression appears in [6] and [, p.94].

2 50 F. M. van Lamoen and P. Yiu Y Y Z Z K P O X X Figure.. Notations We adopt the following notations. For a triangle of sidelengths a, b, c, let S denote twice the area of the triangle, and S = b + c a, S = c + a b, S = a + b c. These satisfy S S + S S + S S = S. More generally, for an arbitrary angle θ, S θ = S cot θ. In particular, S + S + S = a + b + c = S ω, where ω is the rocard angle of triangle. 3. The triangle of medians iven a triangle with sidelengths a, b, c, let m a, m b, m c denote the lengths of the medians. y the pollonius theorem, these are given by m a = 4 (b + c a ), m b = 4 (c + a b ), () m c = 4 (a + b c ). There is a triangle whose sidelengths are m a, m b, m c. See Figure 3. We call this the triangle of medians of. The following useful lemma can be easily established.

3 onstruction of Malfatti squares 5 Lemma. Two applications of the triangle of medians construction gives a similar triangle of similarity factor 3 4. See Figure 3. Figure 3 Figure 3 We present an interesting example of a triangle similar to the triangle of medians which is useful for the construction of the Malfatti triangle. Lemma 3. The pedal triangle of the symmedian point is similar to the triangle of medians, the similarity factor being tan ω. Y Z Z K X Figure 4 Proof. Since S = bcsin, the distance from the centroid to is clearly S 3b. That from the symmedian point K to is Since K and are isogonal conjugates, K = c a + b + c S c = S S ω c. S S ω c = S 3 m a 3bc = bc m a. S 3b ω S ω

4 5 F. M. van Lamoen and P. Yiu This is a diameter of the circle through, K, and its pedals on and. It follows that the distance between the two pedals is bc m a sin = S m a = tan ω m a. S ω S ω From this, it is clear that the pedal triangle is similar to the triangle of medians, the similarity factor being tan ω. Remark. The triangle of medians of has the same rocard angle as. Proposition 4. Let P be a point with pedal triangle XY Z in. The lines through,, perpendicular to the sides Y Z, ZX, XY concur at the isogonal conjugate of P. We shall also make use of the following characterization of the symmedian (Lemoine) point of a triangle. Theorem 5 (Lemoine). The symmedian point is the unique point which is the centroid of its own pedal triangle. 4. Proof of Theorem onsider a triangle with its Malfatti squares. omplete the parallelogram. See Figure 5. Note that triangles X b X c and are congruent. Therefore, is perpendicular to. Note that this line contains the centroid of triangle. Similarly, if we complete parallelograms and, the lines and contain and are perpendicular to and respectively. Y a Z a Y c Z b X b X c Figure 5. onsider as the pedal triangle of in a triangle homothetic to. y Lemoine s theorem, is the symmedian point of. Since

5 onstruction of Malfatti squares 53 is homothetic to, is homothetic to the pedal triangle of the symmedian point K of. In Figure 5, triangle is the image of Y a Z a under the translation by the vector Y a = Z a =. This means that the line is perpendicular to, and to the -side of the pedal triangle of K. Similarly, and are perpendicular to - and -sides of the same pedal triangle. y Proposition 4, the lines,, concur at the isogonal conjugate of K. This means that triangles and are homothetic at the centroid of triangle, and the sides of the Malfatti squares are parallel and perpendicular to the corresponding medians. Denote by λ the homothetic ratio of and. This is also the homothetic ratio of the Malfatti triangle and the pedal triangle of K. In Figure 5, X b + X c = = λa. lso, by Lemmas and 3, X b X c = = λ median of pedal triangle of K = λ tan ω median of triangle of medians of = λ tan ω 3 4 a = 3 λ tan ω a. Since X b + X b X c + X c = X, we have λ( + 3 tan ω) = and λ = + 3tan ω = S ω 3S + S ω. Let h(,λ) be the homothety with center and ratio λ. Since is the symmedian point of, =h(,λ) = λk + ( λ) = 3S + S ω (3S + S ω K). It has homogeneous barycentric coordinates (a + S : b + S : c + S). To compute the coordinates of the vertices of the Malfatti triangle, we make use of the pedals of the symmedian point K on the sidelines. The pedal on is the point X = S ω ((S + S ) + (S + S )). is the point dividing the segment X in the ratio : X = S ω : 3S. = (3S + S ω X) 3S + S ω = (S + (S + S + S ) + (S + S + S )). 3S + S ω For a construction of, see Proposition 6.

6 54 F. M. van Lamoen and P. Yiu Similarly, we have and. In homogeneous barycentric coordinates, these are =(S : S + S + S : S + S + S ), =(S + S + S : S : S + S + S ), =(S + S + S : S + S + S : S). The lines,, intersect the sidelines,, respectively at the points X =(0 : S + S + S : S + S + S ), Y =(S + S + S : 0 : S + S + S ), () Z =(S + S + S : S + S + S : 0). We show that these three intersections are the pedals of a specific point P = (a (S + S) : b (S + S) : c (S + S)). In absolute barycentric coordinates, P = ( (a (S + S) + b (S + S) + c (S + S) ). S(S + S ω ) The infinite point of the perpendiculars to being a + S + S, the perpendicular from P to contains the point = = P + S + S S(S + S ω ) ( a + S + S ) ( (b (S + S) + S (S + S)) + (c (S + S) + S (S + S)) ) S(S + S ω ) (S + S ω ) ((S + S + S ) + (S + S + S )). This is the point X whose homogeneous coordinates are given in () above. Similarly, the pedals of P on the other two lines and are the points Y and Z respectively. These lead to a simple construction of the vertex, as the intersection of the lines X and the line joining to the pedal of P on. This completes the proof of Theorem. Remark. part from,,, the vertices of the Malfatti squares on the sidelines are X b = (0 : 3S + S + S : S + S ), X c = (0 : S + S : 3S + S + S ), Y c = (S + S : 0 : 3S + S + S ), Y a = (3S + S + S : 0 : S + S ), Z a = (3S + S + S : S + S : 0), Z b = (S + S : 3S + S + S : 0).

7 onstruction of Malfatti squares n alternative construction The vertices of the Malfatti triangle are the intersections of the perpendiculars from to the sidelines of triangle with the corresponding lines joining to the pedals of K on the sidelines. simple construction of would lead to the Malfatti triangle easily. Note that divides K in the ratio : K = a + b + c : 3S. On the other hand, the point P is the isogonal conjugate of the Vecten point ( ) V = S + S : S + S :. S + S s such, it can be easily constructed, as the intersection of the perpendiculars from,, to the corresponding sides of the pedal triangles of V. See Figure 6. It is a point on the rocard axis, dividing OK in the ratio OP : PK = a + b + c : S. Y Z K V P O X Figure 6. This leads to a simple construction of the point. Proposition 6. is the intersection of K with HP, where H is the orthocenter of triangle. O P K Figure 7. H Proof. pply Menelaus theorem to triangle OK with transversal HP, noting that OH : H = 3 :. See Figure 7.

8 56 F. M. van Lamoen and P. Yiu 6. Some observations 6.. Malfatti squares not in the interior of given triangle. Sokolowsky [3] mentions the possibility that the Malfatti squares need not be contained in the triangle. Jean-Pierre Ehrmann pointed out that even the Malfatti triangle may have a vertex outside the triangle. Figure 8 shows an example in which both and Y a are outside the triangle. Y a Y c Z b Z a X b X c Figure 8. Proposition 7. t most one of the vertices the Malfatti triangle and at most one of the vertices of the Malfatti squares on the sidelines can be outside the triangle. Y a Z a Y c Z b X b X c Figure 9. Proof. If Y a lies outside triangle, then Z a < π, and Z bz a > π. Since Z a is parallel to, = Z b Z a is obtuse. Under the same hypothesis, if and are on opposite sides of, then Z a < π 4, and > 3π 4. Similarly, if any of Z a, Z b, X b, X c, Y c lies outside the triangle, then correspondingly,,,,, is obtuse. Since at most one of these angles can be obtuse, at most one of the six vertices on the sides and at most one of,, can be outside triangle.

9 onstruction of Malfatti squares locus problem. François Rideau [8] asked, given and, for the locus of for which the Malfatti squares of triangle are in the interior of the triangle. Here is a simple solution. Let M be the midpoint of, P the reflection of in, and Q that of in. onsider the circles with diameters P, M, M, Q, and the perpendiculars l P and l Q to at P and Q. See Figure 0. P M Q l P l Q Figure 0. For an arbitrary point, consider with centroid. (i) is obtuse if is inside the circle with diameter P ; (ii) is obtuse if is inside the circle with diameter M; (iii) is obtuse if is inside the circle with diameter M ; (iv) is obtuse if is inside the circle with diameter Q; (v) is obtuse if is on the side of l P opposite to the circles; (vi) is obtuse if is on the side of l Q opposite to the circles. Therefore, the locus of for which the Malfatti squares of triangle are in the interior of the triangle is the region between the lines l P and l Q with the four disks excised. similar reasoning shows that the locus of for which the vertices,, of the Malfatti triangle of are in the interior of triangle is the shaded region in Figure. P M Q Figure.

10 58 F. M. van Lamoen and P. Yiu 7. eneralization We present a generalization of Theorem in which the Malfatti squares are replaced by rectangles of a specified shape. We say that a rectangle constructed on a side of triangle has shape θ if its center is the apex of the isosceles triangle constructed on that side with base angle θ. We assume 0 < θ < π so that the apex is on the opposite side of the corresponding vertex of the triangle. It is well known that for a given θ, the centers of the three rectangles of shape θ erected on the sides are perspective with at the Kiepert perspector ( K(θ) = S + S θ : The isogonal conjugate of K(θ) is the point S + S θ : S + S θ ). K (θ) = (a (S + S θ ) : b (S + S θ ) : c (S + S θ )) on the rocard axis dividing the segment OK in the ratio tan ω tan θ :. Theorem 8. For a given θ, let (θ) be the intersection of the lines joining (i) the centroid to the pedal of the symmedian point K on, (ii) the vertex to the pedal of K (θ) on. nalogously construct points (θ) and (θ). onstruct rectangles of shape θ on the sides of (θ)(θ)(θ). The remaining vertices of these rectangles lie on the sidelines of triangle. Figure illustrates the case of the isodynamic point J. The Malfatti rectangles Z a Y a, X b Z b and Y c X c have shape π 3, i.e., lengths and widths in the ratio 3 :. Y a Z a K J Y c Z b X b X c Figure. The same reasoning in 6 shows that exactly one of the six vertices on the sidelines is outside the triangle if and only if a median makes an obtuse angle with an adjacent side. If this angle exceeds π + θ, the corresponding vertex of Malfatti triangle is also outside.

11 onstruction of Malfatti squares 59 References [] H. Fukagawa and D. Pedoe, Japanese Temple eometry Problems, harles abbage Research entre, Winnipeg, 989. [] H. Fukagawa and J. F. Rigby, Traditional Japanese Mathematics Problems of the 8th and 9th enturies, ST Press, Singapore, 00. [3] H. Fukagawa and D. Sokolowsky, Problem 03, rux Math., (985) 50; solution, ibid., (986) 9 5, 8 8. [4] R.. Johnson, dvanced Euclidean eometry, 95, Dover reprint, 007. [5]. Kimberling, Triangle centers and central triangles, ongressus Numerantium, 9 (998) 85. [6]. Kimberling, Encyclopedia of Triangle enters, available at [7] F. M. van Lamoen, Friendship among triangle centers, Forum eom., (00) 6. [8] F. Rideau, Hyacinthos message 596, December 8, 007. Floor van Lamoen: Ostrea Lyceum, ergweg 4, 446 N oes, The Netherlands address: fvanlamoen@planet.nl Paul Yiu: Department of Mathematical Sciences, Florida tlantic University, oca Raton, Florida, , US address: yiu@fau.edu

Some Constructions Related to the Kiepert Hyperbola

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