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SOME FAMOUS PROBLEMS AND THEIR SOLUTIONS

Morley’s Theorem, Pick’s Theorem, The Butterfly Problem. Here are three well-known problems, with clear and detailed solutions at the high school level. These proofs are by Gilbert Kessler, and are based on appropriate parts of other more complicated proofs.

[1] Morley’s Theorem. The intersection points of adjacent angle-trisectors of any triangle form an equilateral triangle.

The overall approach will be to use the Law of Sines in ∆AFB to get AF, then use the Law of Cosines in ∆AFE to get FE. Mix well with bits of Geometry, Trigonometry, Identities, and Symmetry, and we’re done! But it’s all clear, detailed, understandable, and only uses high school math.

To start, we will need three small theorems (the first of which is standard), and a lot of straightforward trigonometry.

(1a) If a triangle is inscribed in a circle of diameter d, any side is d times the sine of the opposite angle. (1b) If a triangle is inscribed in a circle of diameter 1, any side is the sine of the opposite angle.

PROOF: Inscribe triangle STU in a circle of diameter d; draw diameter SO, meeting the circle at T', and draw T'U, noting that angles T and T' are equal and angle T'US is a right angle. Then t/d = sin T' = sin T, or t/sin T = d [this approach leads directly to the (Extended) Law of Sines]. Thus t = d sin T = sin T when d = 1. Note that no circle need be pictured for the relationship t = d sin T to hold.

(2) Let a, b, and c be positive angles whose sum is 60°. Later we will need a relationship among the quantities a, 60° + b, and 60° + c, so we will establish that now. Consider a triangle whose angles are a, 60° + b, and 60° + c; note that their sum is 180°. Circumscribe a circle around that triangle, and expand (or contract) it until the diameter of the circle is 1; this does not affect the size of those angles. Then, from (1b), the sides of the triangle are now sin a, sin (60° + b), and sin (60° + c). Applying the Law of Cosines to the triangle, we get sin²a = sin²(60° + b) + sin²(60° + c) - 2 sin (60° + b) sin (60° + c) cos a.

(3) A strange identity: We will show that 4 sin c sin (60° + c) sin (60° - c) = sin 3c

Expanding the left side, it easily becomes 4 sin c (sin²60° cos²c - cos²60° sin²c), and putting in values for the functions of 60° and simplifying, the left side becomes sin c (3 cos2c - sin2c).

Expanding the right side, we get sin 3c = sin (2c + c) = sin 2c cos c + cos 2c sin c = (2 sin c cos c) cos c + (cos2c - sin2c) sin c , which easily becomes sin c (3 cos2c - sin2c). Thus the identity is established.

Now for Morley’s Theorem [See diagram, note that a + b + c must = 60°, and let d be the diameter of the circumscribed circle of ∆ABC].

(4) In ∆ABF, [note that ∠AFB and (a+b) are supplementary]; thus

AF = = {from (1a)} =

        {from (3)}       = 4d sin b sin c sin (60° + c)

(5) Similarly (or by symmetry), we can show that AE = 4d sin b sin c sin (60° + b)

(6) Finally, we apply the Law of Cosines to ∆AFE, getting

     FE2 = AF2 + AE2 - 2•AF•AE cos a . Substituting from the above equations, squaring and factoring, leads to

     FE2 = 16d2 sin2b sin2c [sin2(60° + c) + sin2(60° + b) - 2 sin (60° + c) sin (60° + b) cos a]

            = {from (2)}  16d2 sin2b sin2c sin2a , whereupon

      FE = 4d sin a sin b sin c !

Since this equation is symmetric in a, b, and c, we get the exact same result when finding DF and DE, so ∆DEF is equilateral!

[2] Pick’s Theorem. [In what follows, points will refer to lattice points only, and polygons (including specific ones, such as “rectangle” or “triangle”) will refer to lattice polygons only.]

If a polygon P has b points along its boundary and c points contained within its boundary, then its area AP is given by AP = .

We will use the definitions: A simple polygon is a polygon whose sides do not cross each other, and which contains no “holes.” A lattice polygon is a simple polygon whose vertices are lattice points.

Assumption: Any lattice polygon with more than 3 sides has an internal diagonal. This is obvious for convex polygons, but perhaps not for concave ones.

We will show that Pick’s Theorem works for any [lattice] rectangle whose sides are parallel to the axes, then for any right triangle whose legs are parallel to the axes, then for any triangle, then for any polygon.

(1) We begin by showing that Pick’s Theorem is additive when applied to any two non-overlapping polygons that share a side [that is, if non-overlapping polygons P and Q share a side, and the Theorem applies to P and Q separately, then it applies to P∪Q].

• Any points originally contained in P or Q separately are still contained points in P∪Q.

• Any boundary points other than those on the shared side are still boundary points of P∪Q.

• If there are k lattice points on the shared side, other than its ends, they were boundary points of P and of Q when those areas were considered separately, so they originally contributed [see Theorem] the value of to the area of P and to the area of Q; when looking at P∪Q, these points become contained points, so they contribute [see Theorem] the value of k to the area of P∪Q (equivalent to the total of their original contributions).

• When P and Q were considered separately, the two endpoints of that shared side each originally contributed the value of ½ to the area of P (that totals 1) and ½ to the area of Q (that’s another 1), for a total of 2. These two pairs of points coincide when looking at P∪Q, and now contribute (as two border points) the total value of only ½ + ½ = 1. We seem to have lost 1 from the total area when uniting P and Q. But the Theorem originally subtracted 1 when finding each separate area (for a total of 2 subtracted); now we only subtract 1 when finding the area of P∪Q, so we have gained back that lost 1.

• Thus the area produced by the Theorem when applied to P by itself (AP) + the area produced by the Theorem when applied to Q by itself (AQ) is equal to the area produced by the Theorem when applied to the union of P and Q (AP∪Q). That is, if non-overlapping polygons P and Q share a side, then AP∪Q = AP + AQ . Also notice that given two such adjacent polygons, if the Theorem applies to their union and also applies to one of the polygons, it must apply to the other polygon [simply subtract AP from both sides]. Thus the Theorem is both additive and subtractive.

(2) Consider a rectangle whose sides are parallel to the axes, with two adjacent sides r and s. Our usual formula gives the area as rs. The number of contained points is (r - 1)(s - 1), and the number of boundary points is 2(r - 1) + 2(s - 1) + 4, so the Theorem produces A = = rs. Thus the Theorem applies to any rectangle whose sides are parallel to the axes.

(3) Consider a right triangle whose legs are parallel to the axes, and whose hypotenuse passes through k lattice points other than its ends; if the legs are r and s, our usual area formula gives the area as . Now note that if we were to complete a rectangle of sides r and s, the number of contained points in it would be (r - 1)(s - 1), and this would include k; thus the contained points in the original triangle would be 1/2 of [(r - 1)(s - 1) - k] . The Theorem, applied to the triangle, produces A = = , which equals . Thus the Theorem applies to any right triangle whose legs are parallel to the axes.

(4) Consider any triangle. By enclosing it in a rectangle (as small as possible preferred) whose sides are parallel to the axes, and drawing the proper horizontal and vertical lines from the vertices of the triangle to the sides of the rectangle, we see that the large rectangle is composed of the original triangle plus a number of rectangles and right triangles. By subtracting these small rectangles and right triangles from the large rectangle, one by one, we see that the Theorem must apply to the original triangle. Thus the Theorem applies to any triangle.

(5) Since any quadrilateral can be split into two triangles by drawing an internal diagonal [see our Assumption], the additive property of the Theorem shows that it applies to any quadrilateral. A diagonal splits a pentagon into a quadrilateral and a triangle; a diagonal splits a hexagon into a pentagon and a triangle, or two quadrilaterals. Using an Inductive process, if we assume that the Theorem applies to any polygon of 3, 4, ..., n-1 sides, it must apply to a polygon of n sides by splitting that polygon (with a single internal diagonal) into two polygons of fewer than n sides each and applying the additive property. [Note: If those two polygons have x and y sides respectively, then n = x + y - 2; clearly both x and y must be less than n.] Thus the Theorem applies to any polygon.

NOTE: There is no 3-D analog to Pick’s Theorem for finding the volume of a lattice polyhedron; this can be seen by considering the pyramid with vertices ((0,0,0), (1,0,0), (0,1,0), and (0,0,r) [where r is an integer]. Other than its four vertices, it has no boundary or contained lattice points, but its volume changes as r changes.

[3] The Butterfly Problem.

Given: Chord AMB, with AM ≅ BM, and any chords CME and DMF [Let CF and DE meet AB at P and Q, respectively]

Prove: PM ≅ QM

Draw segments EB, EG (|| BA), GM, GP, GA, GF and mark angles 1, 2, 3, 4.

	[a] Get ∆MAG  ≅ ∆MBE (sas)

AM ≅ BM (given) Arc AG = arc BE (parallel chords intercept equal arcs) Arc AGE = arc BEG (after adding arc GE to each of above) ∠A ≅ ∠B (inscribed angles with equal arcs) AG ≅ BE (equal arcs have equal chords) ∆MAG ≅ ∆MBE (sas)

[b] Get ∆PMG ≅ ∆QME ∠AMG ≅ ∠BME and MG ≅ ME We just need ∠1 congruent to ∠2 to get ∆PMG ≅ ∆QME We see that ∠3 ≅ ∠2, so we must get ∠3 ≅ ∠1

[c] Show that PMGF is cyclic ∠PFG is supp to ∠4 (from cyclic quad CFGE) But ∠4 ≅ ∠BME (from parallel lines) ≅ ∠AMG (see above) Therefore ∠PFG is supplementary to ∠AMG Therefore PMGF is cyclic

Now ∠3 ≅ ∠1 (inscribed angles with same arc) But ∠3 ≅ ∠2 (inscribed angles with same arc), so ∠1 ≅ ∠2

	∆PMG  ≅ ∆QME (asa)

PM ≅ QM!