# Four Points on a Circle Tom Davis October 9, 2002

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1 Four Points on a ircle Tom avis October 9, What is an Interesting Theorem? What makes a Euclidean geometry theorem interesting? There are probably a lot of answers to this, but quite often it is when something occurs that is too lucky to be a coincidence. For example, if you draw some diagram and two of the points of the diagram lie on a line, that is not interesting at all, since every pair of points lie on some line. ut if three points lie on a line when they were not defined to be on that line in the first place, the situation is usually interesting. In fact, if you randomly place three points on a finite portion of a Euclidean plane, selected in a uniform way, the probability that all three will lie on the same line is exactly zero. Similarly, if two non-parallel lines meet at a point, that is not at all interesting, but if three or more do, it may be very interesting. The same sort of idea applies to other figures. If two circles in the plane intersect in two points, or do not intersect at all, this is usually not interesting. ut if they intersect at exactly one point (they are tangent, in other words), the situation is usually very interesting. We begin this document with a short discussion of some tools that are useful concerning four points lying on a circle, and follow that with four problems that can be solved using those techniques. The solutions to those problems are presented at the end of the document. 2 Points on ircles If two points lie on a circle, that is not interesting at all. In fact, for any two Euclidean points, there are an infinite number of circles that pass through both of them. Three points on a circle is also not interesting. Unless the three points happen to lie on the same straight line (which has probability zero of occurring by chance), they lie on a circle. To see this, construct the perpendicular bisectors of the segments joining any two pairs of the points. Since the points do not lie in a line, the perpendicular bisectors are not parallel, and must meet somewhere. ny point on the perpendicular bisector of two points is equidistant from those points, so the point at the intersection of the two perpendicular bisectors is equidistant from all three points. Thus a circle centered 1

2 at this intersection passing through any one of the three points must pass through the other two. Since the first three points define a circle, it is almost impossible for the fourth point to lie on that circle by chance. Thus if, in a diagram, four points that are not in some sense defined to lie on a circle do lie on a circle, that is an interesting occurrence and it may indicate that an interesting theorem can be found. Obviously, if five, six, or more points lie on a circle, that may be even more interesting, and one theorem we will examine here concerns the famous nine point circle where nine points related to any triangle all lie on the same circle. See Section ircle Preliminaries O θ θ θ/2 O θ θ Figure 1: entral ngle and Inscribed ngle There s basically one theorem we need to know to show a large number of interesting things about sets of points on a circle. rcs of circles are measured by the central angle. In the left part of Figure 1, is the center of the circle, and the measure of the arc is the same as the measure of the central angle. oth are called in the figure. The right part of Figure 1 illustrates the theorem. If an angle is inscribed in the circle that intersects the circle in an arc of size, then the measure of the angle is. This is not obvious, and requires proof, but it is proved in every high school geometry course. We must be a little careful here. In the figure, the two points and really determine two arcs in this case a short one that s perhaps in the figure, the rest of the circle (which is about in the figure). If there is any question, we will name the two points that bound the arc in such a way that the arc is the part of the circle from the first 2

3 to the second in a counter-clockwise direction. Thus in the figure, arc is about and arc is about. If, in Figure 1, the point were on the arc (the small one), the angle would measure half of about. This basic fact about angles inscribed in a circle and the fact that a complete circle contains degrees, allows us to prove a wide variety of interesting theorems. d a c b Figure 2: yclic Quadrilateral Since we will be considering the case where four (or more) points lie on a circle, consider some easy (and not so easy) results displayed in Figure 2. In that figure, the four vertices,,, and lie on a circle in order. We will let,,, and represent the lengths of the segments,,, and, respectively. is a quadrilateral that is inscribed in a circle, and is called a cyclic quadrilateral (or sometimes a concyclic quadrilateral). The most interesting and useful result is this: "!" \$#% & (and also, of course, ' (!) *#\$ & ). This is easy to see, since is half the measure of arc (measured counterclockwise) and is half the measure of arc. ut arcs and together make the entire circle, or, so the two opposite angles in any cyclic quadrilateral are supplementary, or in other words, add to make the straight angle. It is not difficult to see the converse; namely, that if two opposite angles in a quadrilateral add to &, then the quadrilateral is cyclic. very important special case of this is when the two opposite angles are right angles, or +. This happens often, and when it does, the resulting quadrilateral is cyclic. 3.1 Ptolemy s Theorem There is a very easy way to determine whether four points lie on a circle if you know the distances between them. onversely, if you know that four points lie on any circle, the formula below, known as Ptolemy s Theorem, holds. 3

4 1 : 1 ; 1 <, 1 - : 0 1 / ; 0 / 0 / In Figure 2, if:, -). /0"12-/3. 0',345, /3. -0 then the four points,,, and all lie on a circle. We will prove one direction of this result in Section rahmagupta s Formula gain, we will not prove this result here since it is not used in the problem set, but it also relates to cyclic quadrilaterals. The usual proof is straightforward, but involves a lot of patience with trigonometric manipulations that begin with the observation that the laws of sines and cosines apply to various triangles in the, - / 0' figure. rahmagupta s formula, -/0 provides a method to calculate 6'7, the area of the cyclic, - / 0 quadrilateral. Here it is: If is a cyclic quadrilateral, whose sides have lengths 9, :, ;, and <, then if = 4 >? is the semiperimeter, its area is given by: 7 9, 6'7 7 = 9 7 = : 7 = ; 7 = < Notice that from rahmagupta s formula it is trivial to deduce Heron s formula for the area of any triangle (remember that, -/ every triangle is concyclic). Here is Heron s formula for the area of a triangle having sides of lengths 9, :, and ; and semiperimeter = 4 >? : 7 9 6'7, = 7 = 9 7 = : 7 =E ; The derivation of Heron s formula is obvious: pick a point near of, but on its circumcircle. Then let approach. The term 7 = < in rahmagupta s formula will tend toward =, - / 0, - /, and the quadrilateral will become more and more like the triangle., -/ 4 Problems 4.1 The Nine Point ircle, F -F /F This is perhaps one of the most amazing theorems in Euclidean geometry. It states that if you begin with any triangle, the nine points consisting of the midpoints of the sides (points,, and in Figure 3), the feet of the altitudes (points G, H, and I ), and the midpoints of the segments from the orthocenter to the vertices of the triangle (points J, K, and L ), all lie on the same circle. (The orthocenter M is the intersection of the three altitudes.) For the proof, see Section 5. 4

5 R E F H Q S G Figure 3: Nine Point ircle 4.2 Miquel s Theorem Figure 4: Miquel s Theorem Figure 4 shows two examples of Miquel s theorem. hoose any triangle and then choose a point on each of its sides. onstruct circles passing through each vertex of the triangle and through the points on the two adjacent sides. ll three of those circles meet at a point. The figure shows that the result seems to hold in two cases, but note that the point of intersection of the three circles may not lie within the original triangle. In fact, the theorem even holds if the points on the sides of the triangle do not lie between the vertices of the triangle. For the proof, see Section Napoleon s Theorem Napoleon s Theorem (see Figure 5) says that given any triangle, if you erect equilateral triangles on the edges of that triangle, all pointing outward, that the centers of those three triangles will form an equilateral triangle. In the figure three circles are drawn to give a hint as to how the proof should proceed. Notice that those three circles all seem to meet at a point. 5

6 F S E R O Q Figure 5: Napoleon s Theorem For the proof, see Section The Simson Line M Figure 6: The Simson Line Let N be any point on the circumcircle of O P QR. From N, drop perpendiculars to each of the sides of the triangle. Show that the points of intersection of the perpendiculars with the sides all lie on a line called the Simson Line. See Figure 6 For the proof, see Section 4.5 Fagnano s Problem Fagnano s problem is the following: Given an acute-angled triangle OSP Q R, find the triangle with the smallest perimeter whose vertices lie on the edges of O P QR. In Figure 7, this triangle with smallest perimeter is OST'U V. The answer is that it is the triangle whose vertices are the feet of the altitudes of the original triangle. The easiest proof uses some interesting arguments involving not only circles, but reflections of the original triangle. For the proof, see Section 9 6

7 F E Figure 7: Fagnano s Problem 5 Proof: Nine Point ircle R E F H N Q S G Figure : Nine Point ircle See Figure. We will prove that all nine points lie on the circle by first showing that the six points W X, Y X, Z, [, \ and ] all lie on a circle. The proof will use the line W Y as the base of the triangle. ut every triangle has three bases, and if we consider the line W ^ to the the base, exactly the same proof would show that the points W X, ^X, _, Z, \, and ` also lie on the same circle. Since these two circles have three points in common: W X, Z, and \, they must be the same circle, and hence all nine points lie on the same circle. Since W X and YX are midpoints of the sides of asw ^ Y, we know that W X Y XbSW Y. Similarly, since Z and [ are midpoints of the sides of asw cdy, Z [3bW Y. ut since both Z [ and W X Y X are parallel to the same line W Y, they are parallel to each other: Z [ebew X Y X. Using the same reasoning, since W X and [ are midpoints of ^ Y and cdy in a ^cdy, W X [fbs^ c. If we look at a ^cdw, repeat the reasoning to show that YX Z*b ^c. 7

8 Since gh i and j h k are both parallel to the same line l m, they are parallel to each other. Thus j h gh i k is a parallelogram. ut it is more than a parallelogram; it is a rectangle. Since l m is part of the altitude of the triangle l mon(j g. ecause j h k and i k are parallel to perpendicular lines, j h k2n5i k. Therefore j h gh i k is a rectangle. Let p be the center of the rectangle. Obviously p is equidistant from j h, g h, i, and k, so all four of those points lie on a circle centered at p. In this circle, j h i and g h k are diameters. Since the lines i q and ksr are altitudes, t j h q i"u t g h r k2u5v w x. ut these right angles subtend the diametersj h i andgh k, so q and r lie on the circle centered at p and passing through points j h, gh, i, and k. Thus we have shown that six of the nine points lie on a circle. s we said above, there is no reason to assume that j g is the only base, and using either j l or l g as base we can show that other sets of six points lie on circles, and since each set has three points in common with the two other sets, all nine points lie on the same circle. 6 Proof: Miquel s Theorem E M F Figure 9: Miquel s Theorem The proof is fairly simple; see Figure 9. Using the labels in Figure 9, begin by considering only two of the circles j q r and lry. They will meet at some point z, and our goal will be to show that the point z also lies on the circle gq y. onstruct lines y'z, q z, and rz, forming two cyclic quadrilaterals j q z)r and ly'z)r t t. Since the quadrilaterals are cyclic, we know that q j r5{ t q z)r)u) } w x and that r l ye{ t y'z)r)u) } w x t. We also know that q j r~{ t r l ye{ t y'g yu } w x t since they are the three angles of a triangle, and we know that q z)r { t t y z)r { q z)y(u5 w x. t pply a little algebra to those equations to conclude that y'g q3{ t q z)y u } w x. If two opposite angles in a quadrilateral add to } w x, we know that it is a cyclic quadri-

9 lateral, so the points ƒ,, and all lie on a circle. Thus the three circles meet at the point. 7 Proof: Napoleon s Theorem F S E R O Q Figure 10: Napoleon s Theorem The proof of Napoleon s Theorem is similar in some ways to the previous proof. See Figure 10. We begin by constructing circles around the three equilateral triangles and we will show that they all meet at a point. In the figure, consider the two circles ˆ and ƒ ˆ. In addition to meeting at ˆ they will meet at another point Š. Since quadrilaterals Š ˆ and ˆŠƒ are cyclic, we know that opposite angles in them add to Œ Ž. Thus Š ˆ) 3 ˆ " ƒš ˆ) " ˆ 'ƒf f Œ Ž. ut the triangles ˆ and Sƒˆ are equilateral so ˆ 5 " ˆ ƒ) 5 Ž. Thus we can conclude that ƒš ˆ( ) Šˆ( ) Ž. Since the angles ƒš ˆ, Š ˆ, and ƒš add to make a full Ž, ƒš 3 ) Ž. ut then since ƒ Š 5 ) ƒ " ( Œ Ž, quadrilateral Šƒ is cyclic, so the three circles surrounding the exterior equilateral triangles meet in a single point Š. Not only that, but the lines Š, ˆŠ, and ƒš all meet at Ž angles. The points,, and are centers both of the equilateral triangles and of the circles, so the lines connecting them are perpendicular to the lines Š, ˆŠ, and ƒš. Thus s Š( " sˆ Š" 3š Ž and since they add to Œ Ž, quadrilateral s Š ˆ is cyclic and sˆ 5 Ž " Œ Ž. Therefore sˆ 3 Ž. ut the same argument can be made to show that ƒ ˆ œ ) ƒ 5 Ž, so is equilateral. 9

10 Proof: The Simson Line The following proof works only if ž lies on the arc between Ÿ and. If it is on one of the other arcs, rename the points appropriately before beginning. M Figure 11: The Simson Line Since the lines dropped from ž are perpendicular to the sides of the triangle, they form right angles, and it is easy to see in Figure 11 that ž) Ÿ, ž) Ÿ and ž) Ÿ are sets of concyclic points. Obviously Ÿž) is also concyclic, since ž is on the circumcircle of Ÿ. Since Ÿ ž) and ž) Ÿ are concyclic sets of points, both Ÿž) and ž)ÿ are supplementary to Ÿ. Therefore they are equal. Since Ÿ ž) " ( Ÿ ž)ÿ Ÿ ž) and ž)ÿ % Ÿ ž)ÿ ( ž)ÿ we find that Ÿ ž) % ž)ÿ. ut since each of those angles lies in a circle, ž)ÿ Ÿ and Ÿ ž)f Ÿ. The equality of the final pair of angles proves that the pointsÿ, and are collinear, since equal vertical angles are guaranteed at. 9 Proof: Fagnano s Problem F H E Figure 12: Equal ngles To prove that the triangle connecting the feet of the altitudes of an acute triangle has the minimum perimeter, we first need to prove a lemma. We will show that this pedal triangle S ' ª in Figure 12 makes equal angles with the bases of the triangles. In other words, we need to show that Ÿ " ) 'ª. 10

11 We begin by considering the two circles passing through «and. It is clear that both pass through, the orthocenter of ±S«² since both quadrilaterals«d and ' d contain two opposite right angles and are hence concyclic. ut ³ d«) ³ d since they are vertical angles, and since the two quadrilaterals «d and d are concyclic, ³ ³ d and ³ «³ d«since they subtend equal arcs of the circles. Thus ³ " )³ «'. X Y Y X Figure 13: Fagnano s Problem With the lemma above, and with a somewhat miraculous construction, the proof of the main theorem is not difficult. See Figure 13. Reflect the original triangle ±S«² across the line ² forming a triangle ± «µ ². Reflect this one across «µ ² forming another triangle ± «µ µ ². ontinue in this way for five reflections, as in the figure. hase angles to show the line µµ «µµ is parallel to «. It is easy to show the reflections of the triangle connecting the feet of the altitudes (the one with solid diamond vertices) forms a straight line after the reflections. ny other triangle (like the one in the figure with dotted diamond vertices) will not be reflected to form a straight path. The beginning and end points of these two paths ( to ~µµ and to µµ ) are equally far apart, since they form a parallelogram. It is a parallelogram since µµ «µµ is parallel to «and the length ~ is the same as the length ~µµ µµ. ut the solid-diamond path is a straight line, so any other path must be longer. Note that the two paths each consist of two copies of the sides of the pedal triangles, so clearly the pedal triangle connecting the bases of the altitudes is the shortest. 11

12 ¼ Ä» Â Â Ã Â À À À Ä Ã 10 Proof: Ptolemy s Theorem ¹)º»¼"½¾¹»3º ¼' 3 5»3º ¹ ¼dÀ The statement of Ptolemy s theorem contains a bunch of products of lengths of segments: (1) Generally, to prove theorems like this we need to convert those products into ratios, and then use similar triangles or some other technique to establish the ratios. It turns out not ¹'Á to» matter much how we start, ¹ but notice» that in equation 1 if we could get the ratio that might help. ut and are not in any pair of similar triangles, ¹ so let s just construct a line» that creates a pair of similar triangles with the side in one corresponding to side in the other. H ¼» Figure 14: Ptolemy s Theorem ¹ In Figure 14, construct the line from outside the cyclic quadrilateral that makes the same angle with that does with ¹ ¼* and» intersects the line at Â. Now» ¼ we have the ¹» similarity ¼' that we want. Ã Â since both are ¹ ¼ equal to Ã»¼ plus Ã or Ã Â which were constructed to be equal. Since Ã and Ã are inscribed ¹ ¼Å the same» circle, they are also equal, so by angle-angle similarity, we know that Ä Â. ¹»»¼ ut there s another pair of similar triangles. Since Ã and Ã are opposite»¼ angles in¼ a cyclic quadrilateral, they are ¹»f supplementary. ¼ It is also obvious ¼% that» ¹ Ã and Ã Â are supplementary, so Ã Ã Â. Since ¹»)Å Ã Â ¼ (by construction), we have, again by angle-angle similarity, that Ä Â. Finally, notice that» e» ¼"½2¼ (2)» ¹» ¹ ¼ÇÆ È ÉsÊ Ë Ë Í Ê Ì Ì Î»3º ¹¼ ¹ 5»¼"½ ¼"º ¹» ¹ ¹ From the similarities of the pairs of triangles, we have: If we solve for obtain: Â and When we multiply through by Â in equations 3 and substitute them into equation 2, we we obtain Ptolemy s theorem.» ¼ (3) 12

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