A Fine Use of Transformations

Transcription

1 ine Use of ransformations ithvik Pasumarty Wayzata High chool, Plymouth, Minnesota, United tates 1 Introduction In this paper, we discuss a very fundamental, yet underexposed, idea to geometry. n affine transformation between two affine spaces U, V is a transformation ϕ : U V such that for any three collinear points,, and, the points = ϕ(), = ϕ(), and = ϕ() are collinear, and = In the scope of this paper, we can think of an affine transformation of the plane we will work only with affine transformations between and itself to be any transformation that preserves collinearity and ratios of segments. heorem 1.1. ll affine transformations can be written as ϕ(x, y) = (a 1 x + b 1 y + c 1, a x + b y + c ). Proof. uppose ϕ is an affine transformation, and let ψ(p ) = ϕ(p ) ϕ(0). hen 0,, and α are mapped to ϕ(0), ϕ(), and ϕ(α) = ϕ(0) + α(φ() φ(0)), so ψ(α) = αψ(). lso, ψ( + ) = ψ ( ) + = ( ψ ( ) ( + ψ )) = ψ() + ψ(), so ψ is a linear map, as desired. orollary. We can classify all affine transformations as a combination rotations, translations, dilations, and shears. shear is a transformation of the form (x, y) (x + ky, y). he most important property of affine transformations is that they preserve ratios of lengths and areas. his often allows us to make very magical assumptions about a configuration, like that a triangle is equilateral, a parallelogram is a square, or an ellipse is a circle. ome pplications We begin with a problem of lines. xample.1 (Putnam 001). has an area 1. Points,, lie, respectively, on sides,, such that bisects at point, bisects at point, and bisects at point. ind the area of. olution. When amuel Zbarsky presented his solution to this problem at the arnegie Mellon Putnam eminar, he began it with, We assume that is both equilateral and isosceles-right. hat is, a shear along will preserve all of the ratios in the problem, so we swap back and forth between whichever one makes calculations easiest. onsider the following three diagrams: 1

2 y symmetry in the equilateral case, we can see that = = = r. In the isosceles-right case, we can use the fact that bisects to obtain the identity (1 r) ( 1 1r) = 1 = r = 3 we know that = 1 (1 r), so we can compute = r. lso, since =, we have 1 r. Now = = 1 = 1 = 1 r + (1 r) 1 + (1 r) = r inally, we have [ ] = [] = 1 r r [] = 1 r [] = 7 3 [] ffine transformations are excellent for problems involving ellipses, since an ellipse is the image of a circle under an affine transformation. xample. (Lehigh 01). What is the area of the largest ellipse that can be inscribed in a triangle with sides 3,, and? olution. We can perform an affine transformation mapping the 3 triangle to an equilateral triangle: he ellipse of maximal area in the original triangle will map to the ellipse of maximal area in the new triangle, which will be its incircle by symmetry. In the equilateral triangle, we compute the ratio of the incircle to total area: πr K = π(k/s) = πk = π(l 3/) = π K s (3l/) 3 3 ince the transformation preserves ratios of areas, we just need to scale back to get the desired ellipse s π area: 3 1 (3)() = π 3 3. his ellipse of maximal area is called the teiner inellipse, and its center is the centroid of the triangle. 3 ome More ems ry the following problems using the right affine transformations. xercise 3.1 (Putnam 199). Let be the area of the region in the first quadrant bounded by the line y = 1 x, the x-axis, and the ellipse 1 9 x + y = 1. ind the positive number m such that is equal to the area in the first quadrant bounded by the line y = mx, the y-axis, and the ellipse 1 9 x + y. xercise 3. (Morocco 01). line passes through parallelogram and intersects the segments,, and at,, and, respectively. Prove that + =.

3 xercise 3.3. In let be the foot of the perpendicular from to, and let X be any point on. X and X intersect and at and, respectively. how that X = X. xercise 3. (IM 01). block of wood has the shape of a right circular cylinder with radius 6 and height 8, and its entire surface has been painted blue. Points and are chosen on the edge of one of the circular faces of the cylinder so that  on that face measures. he block is then sliced in half along the plane that passes through point, point, and the center of the cylinder, revealing a flat, unpainted face on each half. he area of one of these unpainted faces is a π + b c, where a, b, and c are integers and c is not divisible by the square of any prime. ind a + b + c. xercise 3. (MIM 016). Let P be the unique parabola in the xy-plane which is tangent to the x-axis at (, 0) and to the y-axis at (0, 1). We say a line l is P-friendly if the x-axis, y-axis, and P divide l into three segments, each of which has equal length. If the sum of the slopes of all P-friendly lines can be written in the form m for m and n positive relatively prime integers, find m + n. n olutions xercise 3.1 (Putnam 199). Let be the area of the region in the first quadrant bounded by the line y = 1 x, the x-axis, and the ellipse 1 9 x + y = 1. ind the positive number m such that is equal to the area in the first quadrant bounded by the line y = mx, the y-axis, and the ellipse 1 9 x + y. olution. We could solve this problem using a nasty integral. Instead, we scale along the x-axis, transforming the ellipse into a circle: he y = 1x line is transformed into the line y = 3 x. y symmetry, the y = mx line must be transformed into the line y = x. ransforming back, we get m =. 3 9 xercise 3. (Morocco 01). line passes through parallelogram and intersects the segments,, and at,, and, respectively. Prove that + =. olution. Perform an affine transformation mapping the parallelogram to the square [0, 1]. Now we can let = (u, 0) and = (0, v) to get = ( uv, uv u+v u+v). We conclude with the relation = u+v u v uv. xercise 3.3. In let be the foot of the perpendicular from to, and let X be any point on. X and X intersect and at and, respectively. how that X = X. 3

4 olution. Perform a dilation centered at along the axis so that X is the orthocenter of. Note that this would preserve the fact that and are reflections over. X X ince X and X are cyclic, X = X = 90 = X = X. xercise 3. (IM 01). block of wood has the shape of a right circular cylinder with radius 6 and height 8, and its entire surface has been painted blue. Points and are chosen on the edge of one of the circular faces of the cylinder so that  on that face measures. he block is then sliced in half along the plane that passes through point, point, and the center of the cylinder, revealing a flat, unpainted face on each half. he area of one of these unpainted faces is a π + b c, where a, b, and c are integers and c is not divisible by the square of any prime. ind a + b + c. olution. he trace of the cut is a truncated ellipse. he length of one axis is the diameter of the circle, 1. We can calculate the length of the other axis from similar triangles to be 0. urthermore, it is truncated at a distance from the center. It remains to use this information to calculate the area of this truncated ellipse. We perform an affine transformation mapping the ellipse to a truncated circle of radius 6. hen this circle is truncated at distance 6 = 3 from the center, so the two arcs of the truncated circle each have measure 60. his means the area of the truncated circle is 6 π caling back, that is 6 (1π ) = 0π = 1π xercise 3. (MIM 016). Let P be the unique parabola in the xy-plane which is tangent to the x-axis at (, 0) and to the y-axis at (0, 1). We say a line l is P-friendly if the x-axis, y-axis, and P divide l into three segments, each of which has equal length. If the sum of the slopes of all P-friendly lines can be written in the form m for m and n positive relatively prime integers, find m + n. n olution. cale along the y-axis so that (0, 1) is replaced with (0, ). he answer will be 1 times as big as this new problem s answer. otate the parabola, and using the angle addition formulas, we find that for each slope m in this new frame, the slope in the original frame is m 1. m+1

5 We are now looking at the parabola that is tangent to the lines y = ±x at ( ±, ). imple calculations reveal this to be the parabola P : y = x +. Let = (a, a) and = (b, b), and we want ( 1 a + b, 1a + b) and ( a + 1b, a + 1b) to be on P : { 1a + b = ( 1 a + b) a + 1b = ( a + 1b) ubtracting, we get a + b = (a + b)(b a). a + b = 0 means that the line is horizontal, which gives way to two solutions easily verifiable from the intermediate value theorem. Otherwise, b a = = m = b+a = b 1. hus, in the original frame the slope is 1. We can use Vieta s formulas in the b a b original equation, 1 3 (b ) + 3 b = ( 1 3 (b ) + ) 3 b + to evaluate the sum of these two slopes to be 36. dding the two horizontal lines that each contribute 1 to the sum, we get that the sum of all of the slopes is 36, which scales back to 36 1 = 3. ithvik Pasumarty Wayzata High chool Plymouth, Minnesota, United tates rithvik.pasumarty@gmail.com