Advanced High-School Mathematics

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Advanced High-School Mathematics David B. Surowski Shanghai American School Singapore American School January 29, 2011Chapter 1 Advanced Euclidean Geometry 1.1 Role of Euclidean Geometry in High-School Mathematics If only because in one's \further" studies of mathematics, the results (i.e., theorems) of Euclidean geometry appear only infrequently, this subject has come under frequent scrutiny, especially over the past 50 years, and at various stages its very inclusion in a high-school mathe- matics curriculum has even been challenged. However, as long as we continue to regard as important the development of logical, deductive reasoning in high-school students, then Euclidean geometry provides as e ective a vehicle as any in bringing forth this worthy objective. The lofty position ascribed to deductive reasoning goes back to at least the Greeks, with Aristotle having laid down the basic foundations of such reasoning back in the 4th century B.C. At about this time Greek geometry started to ourish, and reached its zenith with the 13 books of Euclid. From this point forward, geometry (and arithmetic) was an obligatory component of one's education and served as a paradigm for deductive reasoning. A well-known (but not well enough known) anecdote describes for- mer U.S. president Abraham Lincoln who, as a member of Congress, had nearly mastered the rst six books of Euclid. By his own admis- sion this was not a statement of any particular passion for geometry, but that such mastery gave him a decided edge over his counterparts is dialects and logical discourse. Lincoln was not the only U.S. president to have given serious thought 12 CHAPTER 1 Advanced Euclidean Geometry to Euclidean geometry. President James Gar eld published a novel proof in 1876 of the Pythagorean theorem (see Exercise 3 on page 4). As for the subject itself, it is my personal feeling that the logical arguments which connect the various theorems of geometry are every bit as fascinating as the theorems themselves So let's get on with it ... 1.2 Triangle Geometry 1.2.1 Basic notations We shall gather together a few notational conventions and be reminded of a few simple results. Some of the notation is as follows: A; B; C labels of points AB The line segment joining A and B AB The length of the segment AB (AB) The line containing A and B c A The angle at A c CAB The angle between CA and AB 4ABC The triangle with vertices A; B, and C 0 0 0 0 0 0  4ABC 4ABC The triangles4ABC and4ABC are congruent = 0 0 0 0 0 0 4ABC4ABC The triangles4ABC and4ABC are similarSECTION 1.2 Triangle Geometry 3 1.2.2 The Pythagorean theorem One of the most fundamen- tal results is the well-known Pythagorean Theorem. This 2 2 2 states thata +b =c in a right triangle with sides a and b and hypotenuse c. The gure to the right indicates one of the many known proofs of this fundamental result. Indeed, the area of the 2 \big" square is (a +b) and can be decomposed into the area of the smaller square plus the areas of the four congruent triangles. That is, 2 2 (a +b) =c + 2ab; 2 2 2 which immediately reduces to a +b =c . Next, we recall the equally well- known result that the sum of the  interior angles of a triangle is 180 . The proof is easily inferred from the diagram to the right. Exercises 1. Prove Euclid's Theorem for Proportional Segments, i.e., given the right triangle4ABC as indicated, then 2 2 2 h =pq; a =pc; b =qc: 2. Prove that the sum of the interior angles of a quadrilateralABCD  is 360 .4 CHAPTER 1 Advanced Euclidean Geometry 3. In the diagram to the right,4ABC is a right triangle, segments AB and AF are perpendicular and equal in length, and EF is per- pendicular to CE. Set a = BC; b = AB; c = AB, and de- 1 duce President Gar eld's proof of the Pythagorean theorem by com- puting the area of the trapezoid BCEF . 1.2.3 Similarity In what follows, we'll see that manyif not mostof our results shall rely on the proportionality of sides in similar triangles. A convenient statement is as follows. B Similarity. Given the similar tri- 0 0 angles4ABC4ABC , we have A' that C' 0 0 0 0 AB BC AC = = : AB BC AC A C Conversely, if 0 0 0 0 AB BC AC = = ; AB BC AC 0 0 then triangles4ABC4ABC are similar. 1 James Abram Gar eld (18311881) published this proof in 1876 in theJournalofEducation (Volume 3 Issue 161) while a member of the House of Representatives. He was assasinated in 1881 by Charles Julius Guiteau. As an aside, notice that Gar eld's diagram also provides a simple proof of the fact that perpendicular lines in the planes have slopes which are negative reciprocals.SECTION 1.2 Triangle Geometry 5 0 0 Proof. Note rst that 4AAC 0 0 and4CAC clearly have the same 0 areas, which implies that4ABC 0 and 4CAB have the same area (being the previous common area plus the area of the common trian- 0 0 gle4ABC ). Therefore 1 0 0 AB hAB 2 = 1 AB hAB 2 0 0 area4ABC = 0 area4ABC 0 0 area4ABC = 0 area4CAB 1 0 0 hBC 2 = 1 0 hBC 2 0 BC = BC 0 0 0 AB AC In an entirely similar fashion one can prove that = : AB AC Conversely, assume that B 0 0 AB BC = : AB BC In the gure to the right, the point A' 00 C has been located so that the seg- C" 0 00 ment AC is parallel to AC. But 0 00 then triangles4ABC and4ABC C' are similar, and so A 00 0 0 BC AB BC = = ; C BC AB BC 00 0 0 00 0 0 i.e., thatBC =BC . This clearly implies thatC =C , and so AC is parallel to AC. From this it immediately follows that triangles6 CHAPTER 1 Advanced Euclidean Geometry 0 0 4ABC and4ABC are similar. Exercises 0 0 0 c 0d 0 0 1. Let 4ABC and 4ABC be given with ABC = ABC and 0 0 0 0 AB BC 0 0 0 = . Then4ABC4ABC . AB BC A 2. In the gure to the right, AD =rAB; AE =sAC. D Show that E Area4ADE =rs: Area4ABC C B 3. Let4ABC be a given triangle and let Y; Z be the midpoints of AC; AB, respectively. Show that (XY ) is parallel with (AB). (This simple result is sometimes called the Midpoint Theorem) B 4. In4ABC, you are given that AY CX BX 1 Z X = = = ; YC XB ZA x C where x is a positive real number. Assuming that the area of4ABC is 1, compute the area of4XYZ as Y a function of x. A 5. LetABCD be a quadrilateral and letEFGH be the quadrilateral formed by connecting the midpoints of the sides ofABCD. Prove that EFGH is a parallelogram.SECTION 1.2 Triangle Geometry 7 6. In the gure to the right,ABCD is a parallelogram, and E is a point on the segment AD. The point F is the intersection of lines (BE) and (CD). Prove thatABFB = CFBE: 7. In the gure to the right, tangents to the circle atB andC meet at the point A. A point P is located on รน the minor arc BC and the tangent to the circle at P meets the lines (AB) and (AC) at the pointsD and c E, respectively. Prove thatDOE = 1 c BOC, whereO is the center of the 2 given circle. 1.2.4 \Sensed" magnitudes; The Ceva and Menelaus theo- rems In this subsection it will be convenient to consider the magnitudeAB of 2 the line segment AB as \sensed," meaning that we shall regard AB as being either positive or negative and having absolute value equal to the usual magnitude of the line segment AB. The only requirement that we place on the signed magnitudes is that if the points A; B; and C are colinear, then 8 0 if AB and BC are in the same direction ABBC = : 0 if AB and BC are in opposite directions. 2 IB uses the language \sensed" rather than the more customary \signed."8 CHAPTER 1 Advanced Euclidean Geometry This implies in particular that for signed magnitudes, AB = 1: BA Before proceeding further, the reader should pay special attention to the ubiquity of \dropping altitudes" as an auxiliary construction. Both of the theorems of this subsec- tion are concerned with the following con guration: we are given the trian- gle4ABC and points X; Y; and Z on the lines (BC); (AC); and (AB), respec- tively. Ceva's Theorem is concerned with the concurrency of the lines (AX); (BY ); and (CZ). Menelaus' Theorem is con- cerned with the colinearity of the points X; Y; andZ. Therefore we may regard these theorems as being \dual" to each other. In each case, the relevant quantity to consider shall be the product AZ BX CY   ZB XC YA Note that each of the factors above is nonnegative precisely when the points X; Y; and Z lie on the segments BC; AC; and AB; respec- tively. The proof of Ceva's theorem will be greatly facilitated by the fol- lowing lemma:SECTION 1.2 Triangle Geometry 9 Lemma. Given the triangle 4ABC, letX be the intersection of a line throughA and meeting (BC). LetP be any other point on (AX). Then area4APB BX = : area4APC CX Proof. In the diagram to the right, altitudes BR and CS have been constructed. From this, we see that 1 area4APB APBR 2 = 1 area4APC APCS 2 BR = CS BX = ; CX where the last equality follows from the obvious similarity 4BRX4CSX: Note that the above proof doesn't depend on where the line (AP ) in- tersects (BC), nor does it depend on the position of P relative to the line (BC), i.e., it can be on either side. Ceva's Theorem. Given the triangle4ABC, lines (usually called Cevians are drawn from the vertices A,B, andC, withX,Y , andZ, being the points of intersections with the lines (BC); (AC), and (AB), respectively. Then (AX); (BY ); and (CZ) are concurrent if and only if AZ BX CY   = +1: ZB XC YA10 CHAPTER 1 Advanced Euclidean Geometry Proof. Assume that the lines in question are concurrent, meeting in the point P . We then have, applying the above lemma three times, that area4APC area4APB area4BPC 1 =   area4BPC area4APC area4BPA AZ BX CY =   : ZB XC YA . To prove the converse we need to prove that the lines (AX); (BY ); and (CZ) are concurrent, given that AZ BX CY   = 1: ZB XC YZ 0 Let Q = (AX)\ (BY ); Z = (CQ)\ (AB): Then (AX); (BY ), 0 and (CZ ) are concurrent and so 0 AZ BX CY   = 1; 0 ZB XC YZ which forces 0 AZ AZ = : 0 ZB ZB 0 This clearly implies thatZ =Z , proving that the original lines (AX); (BY ); and (CZ) are concurrent. Menelaus' theorem is a dual version of Ceva's theorem and concerns not lines (i.e., Cevians) but rather points on the (extended) edges ofSECTION 1.2 Triangle Geometry 11 the triangle. When these three points are collinear, the line formed is called a transversal. The reader can quickly convince herself that there are two con gurations related to4ABC: As with Ceva's theorem, the relevant quantity is the product of the sensed ratios: AZ BX CY   ; ZB XC YA in this case, however, we see that either one or three of the ratios must be negative, corresponding to the two gures given above. Menelaus' Theorem. Given the triangle4ABC and given points X; Y; and Z on the lines (BC); (AC); and (AB), respectively, then X; Y; and Z are collinear if and only if AZ BX CY   =1: ZB XC YA Proof. As indicated above, there are two cases to consider. The rst case is that in which two of the pointsX; Y; orZ are on the triangle's sides, and the second is that in which none of X; Y; or Z are on the triangle's sides. The proofs of these cases are formally identical, but for clarity's sake we consider them separately.12 CHAPTER 1 Advanced Euclidean Geometry Case 1. We assume rst that X; Y; andZ are collinear and drop altitudesh ; h ; andh as indicated 1 2 3 in the gure to the right. Using ob- vious similar triangles, we get AZ h BX h CY h 1 2 3 = + ; = + ; = ; ZB h XC h YA h 2 3 1 in which case we clearly obtain AZ BX CY   =1: ZB XC YA To prove the converse, we may assume that X is on BC, Z is on AZ BX CY 0 AB, and thatY is on (AC) with   =1: We letX be the ZB XC YA intersection of (ZY ) with BC and infer from the above that 0 AZ BX CY   =1: 0 ZB XC YA 0 BX BX 0 It follows that = , from which we infer easily thatX =X; and 0 XC X C so X; Y; and Z are collinear. Case 2. Again, we drop altitudes from A, B, and C and use obvious similar tri- angles, to get AZ h BX h AY h 1 2 1 = ; = ; = ; ZB h XC h YC h 2 3 3 it follows immediately that AZ BX CY   =1: ZB XC YA The converse is proved exactly as above.SECTION 1.2 Triangle Geometry 13 1.2.5 Consequences of the Ceva and Menelaus theorems As one typically learns in an elementary geometry class, there are sev- eral notions of \center" of a triangle. We shall review them here and show their relationships to Ceva's Theorem. Centroid. In the triangle4ABC lines (AX); (BY ); and (CZ) are drawn so that (AX) bisects BC, (BY ) bisects CA, and (CZ) bisects AB That the lines (AX); (BY ); and (CZ) are con- current immediately follows from Ceva's Theorem as one has that AZ BX CY   = 1 1 1 = 1: ZB XC YZ The point of concurrency is called the centroid of4ABC. The three Cevians in this case are called medians. Next, note that if we apply the Menelaus' theorem to the triangle 4ACX and the transversal de ned by the pointsB; Y and the centroid P , then we have that AY CB XP 1 =   ) YC BX PA XP XP 1 1 = 1 2 ) = : PA PA 2 Therefore, we see that the distance of a triangle's vertex to the centroid is exactly 1/3 the length of the corresponding median.14 CHAPTER 1 Advanced Euclidean Geometry Orthocenter. In the trian- gle4ABC lines (AX); (BY ); and (CZ) are drawn so that (AX) ? (BC), (BY )? (CA), and (CZ)? (AB). Clearly we either have AZ BX CY ; ; 0 ZB XC YA or that exactly one of these ratios is positive. We have AZ CZ 4ABY4ACZ) = : AY BY Likewise, we have BX AX 4ABX4CBZ) = and 4CBY4CAX BZ CZ CY BY ) = : CX AX Therefore, AZ BX CY AZ BX CY CZ AX BY   =   =   = 1: ZB XC YA AY BZ CX BY CZ AX By Ceva's theorem the lines (AX); (BY ); and (CZ) are concurrent, and the point of concurrency is called the orthocenter of4ABC. (The line segments AX; BY ; and CZ are the altitudes of4ABC.) Incenter. In the triangle4ABC lines (AX); (BY ); and (CZ) are drawn so c that (AX) bisects BAC, (BY ) bisects c c ABC, and (CZ) bisects BCA As we show below, that the lines (AX); (BY ); and (CZ) are concurrent; the point of concurrency is called the incenter of 4ABC. (A very interesting \extremal"SECTION 1.2 Triangle Geometry 15 property of the incenter will be given in Exercise 12 on page 153.) However, we shall proceed below to give another proof of this fact, based on Ceva's Theorem. Proof that the angle bisectors of 4ABC are concurrent. In order to accomplish this, we shall rst prove the Angle Bisector Theorem. We are given the triangle4ABC with line segment BP (as indicated to the right). Then AB AP c c = ,ABP =PBC: BC PC Proof ((). We drop altitudes fromP to (AB) and (BC); call the points so determined Z and Y , re- spectively. Drop an altitude from B to (AC) and call the resulting point X. Clearly PZ = PY as  4PZB =4PYB. Next, we have AB BX BX 4ABX4APZ) = = : AP PZ PY Likewise, CB BX 4CBX4CPY) = : CP PY Therefore, AB APBX PY AP =  = : BC PY CPBX CP16 CHAPTER 1 Advanced Euclidean Geometry AB AP ()). Here we're given that = : Let BC PC 0 P be the point determined by the angle 0 c bisector (BP ) of ABC. Then by what has already been proved above, we have 0 AP AP = : But this implies that 0 BC P C 0 AP AP 0 = )P =P : 0 PC P C Conclusion of the proof that angle bisectors are concurrent. First of all, it is clear that the relevant ratios are all positive. By the Angle Bisector Theorem, AB AY BC BZ AB BX = ; = ; = ; BC YC CA ZA AC XC therefore, AZ BX CY CA AB BC   =   = 1: BZ XC YA BC AC AB Ceva's theorem now nishes the job Exercises 1. The Angle Bisector Theorem involved the bisection of one of the given triangle's interior angles. Now let P be a point on the line (AC) external to the segment AC. Show that the line (BP ) bisects the external angle at B if and only if AB AP = : BC PC 2. You are given the triangle4ABC. Let X be the point of inter- c section of the bisector of BAC with BC and let Y be the point c of intersection of the bisector ofCBA with AC. Finally, letZ be the point of intersection of the exterior angle bisector at C with 3 the line (AB). Show that X; Y; and Z are colinear. 3 What happens if the exterior angle bisector at C is parallel with (AB)?SECTION 1.2 Triangle Geometry 17 3. Given4ABC and assume that X is on (BC), Y is on (AC) and Z is on (AB). Assume that the Cevians (AX) (BY ); and (CZ) are concurrent, meeting at the point P . Show that PX PY PZ + + = 1: AX BY CZ 4. Given the triangle4ABC with incenterP , prove that there exists a circleC (called the incircle of4ABC) with center P which is inscribed in the triangle4ABC: The radius r of the incircle is often called the inradius of4ABC. 5. Let4ABC have side lengths a = BC; b = AC; and c = AB, and let r be the inradius. Show that the area of4ABC is equal r(a+b+c) to . (Hint: the incenter partitions the triangle into three 2 smaller triangles; compute the areas of each of these.) 6. Given the triangle4ABC. Show that the bisector of the internal angle bisector at A and the bisectors of the external angles at B and C are concurrent. 7. Given4ABC and pointsX; Y; and Z B Z in the plane such that X \ABZ = \CBX; \BCX = \ACY; A \BAZ = \CAY: C Show that (AX); (BY ), and (CZ) Y are concurrent. 8. There is another notion of \center" of the triangle4ABC. Namely, construct the linesl ; l ; andl so as to be perpendicular bisectors 1 2 3 of AB; BC, and CA, respectively. After noting that Ceva's theorem doesn't apply to this situation, prove directly that the lines l ; l ; and l are concurrent. The point of concurrency is 1 2 3 called the circumcenter of4ABC. (Hint: argue that the point of concurrency of two of the perpendicular bisectors is equidistant to all three of the vertices.) If P is the circumcenter, then the common value AP = BP = CP is called the circumradius18 CHAPTER 1 Advanced Euclidean Geometry of the triangle4ABC. (This is because the circumscribed circle containing A; B; and C will have radius AP .) 9.4ABC has side lengths AB = 21; AC = 22; and BC = 20: Points D and E are on sides AB and AC, respectively such that DEk BC and DE passes through the incenter of4ABC. Compute DE. B 10. Here's another proof of Ceva's the- orem. You are given4ABC and concurrent Cevians AX; BY ; N and CZ, meeting at the point P . M Z X Construct the line segments AN P and CM, both parallel to the Ce- vian BY . Use similar triangles to C conclude that Y AY AN CX CM BZ BP A = ; = ; = ; YC CM XB BP ZA AN AZ BX CY and hence that   = 1: ZB XC YA 11. Through the vertices of the triangle4PQR lines are drawn lines which are parallel to the opposite sides of the triangle. Call the new triangle4ABC. Prove that these two triangles have the same centroid. 12. Given the triangle4ABC, letC be the inscribed circle, as in Exercise 4, above. Let X; Y; and Z be the points of tangency ofC (on the sides BC; AC; AB, respectively) and show that the lines (AX); (BY ); and (CZ) are concurrent. The point of concurrency is called the Gergonne point of the circleC. (This is very easy once you note that AZ =YZ, etc.) 13. In the gure to the right, the dotted segments represent angle bisectors. Show that the points P; R; and Q are colinear.SECTION 1.2 Triangle Geometry 19 14. In the gure to the right, three cir- cies of the same radius and centers X; Y and Z are shown intersecting at points A; B; C, and D, with D the common point of intersection of all three circles. Show that (a) D is the circumcenter of 4XYZ, and that (b)D is the orthocenter of4ABC. (Hint: note that YZCD is a rhombus.) 15. Show that the three medians of a triangle divide the triangle into six triangle of equal area. 0 16. Let the triangle4ABC be given, and let A be the midpoint of 0 0 BC,B the midpoint of AC and letC be the midpoint of AB. Prove that 0 0 0 (i)4ABC 4ABC and that the ratios of the corresponding sides are 1:2. 0 0 0 (ii)4ABC and4ABC have the same centroid. 0 0 0 (iii) The four triangles determined within4ABC by4ABC are all congruent. 0 0 0 (iv) The circumcenter of4ABC is the orthocenter of4ABC . 0 0 0 The triangle4ABC of4ABC formed above is called the me- dial triangle of4ABC. 17. The gure below depicts a hexagram \inscribed" in two lines. Us- ing the prompts given, show that the lines X; Y; andZ are colin- ear. This result is usually referred to Pappus' theorem.

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