Distance from point to midpoint - August 5-9, 1996
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The lengths of the sides of a triangle are 25, 29, and 36. There is a point on the longest side of the triangle whose distance from the opposite vertex is 20. What is the distance from this point to the midpoint of the shortest side?
This week's problem took almost as much effort to describe coherently as to solve. Look especially for the solutions that noted the existence of two right triangles. Here are the best explanations:Thomas S. Kuo School: Murray Middle School, Ridgecrest, California Grade: 8th A * * * * * x * * * y * E * * * * * * * * D * * * * * * * * * * * * B * * * * * * * * * * * * * * * * * * * * * C AB = 25, BC = 29, AC = 36 BD = 20, AE = EB = 12.5 AD = x, ED = y The length of DE is 12.5 by using the law of cosines. (1) In triangle ABC BC^2 = AB^2 + AC^2 - 2*AB*AC*cos(A) 29^2 = 25^2 + 36^2 - 2*29*25*cos(A) cos(A) = 0.6 (2) In triangle ABD BD^2 = AB^2 + AD^2 - 2*AB*AD*cos(A) 20^2 = 25^2 + x^2 - 2*25*x*0.6 x^2 - 30*x + 225 = 0 = (x - 15)^2 x = 15 (3) In triangle ABD DE^2 = AE^2 + AD^2 - 2*AE*AD*cos(A) y^2 = 12.5^2 + x^2 - 2*12.5*x*cos(A) y^2 = 12.5^2 + 15^2 - 2*12.5*15*0.6 y^2 = 12.5^2 y = 12.5
Chui Yuk Man Grade: Secondary 6 School: Queen Elizabeth School Let the triangle be ABC where AB=25, BC=36 and AC=29 Let D be the point on BD such that AD=20 E is the mid-point on AB and we want to know the length of DE By using Area = [s(s-a)(s-b)(s-c)]^0.5 where a, b and c are the lengths of a triangle and s=(a+b+c)/2, Area of the triangle = 360. So AD is perpendicular to BC (since 20x36/2=360 and if BC is the base, AD must be the height). Therefore triangle ABD is a right angled triangle and if it is inscribed in a circle, AB must be the diameter and the radius is 25/2 = 12.5 Since DE is also the radius, so DE = 12.5
Justin Lam Grade: 7th/8th School: Sequoia Middle School Given: Triangle ABC AB=25 BC=29 AC=36 D is on AC such that BD=20 E is the midpoint of AB Find: The length of DE Using Laws of Cosines, (BC)^2=(AB)^2+(AC)^2-2(AB)(AC)cos(A) (on Triangle ABC) (29)^2=(25)^2+(36)^2-2(25)(36)cos(A) cos(A) = (-1080)/(-1800) = 0.6 Using Laws of Cosines again on Triangle ABD, (BD)^2=(AB)^2+(AD)^2-2(AB)(AD)cos(A) (20)^2=(25)^2+(AD)^2-2(25)(AD)(0.6) (AD)^2-30(AD)+225 = 0 (AD-15)^2 = 0 AD = 15 Since E is the midpoint of AB, AE = 25/2. Use Laws of Cosines one more time on Triangle ADE, (DE)^2=(AE)^2+(AD)^2-2(AE)(AD)cos(A) (DE)^2=(25/2)^2+(15)^2-2(25/2)(15)(0.6) (DE)^2=625/4 DE = 25/2. Note: As it turns out, Triangle ABD is a 15-20-25 right triangle and Triangle BCD is a 20-29-36 right triangle.
Ken Duisenberg Stanford '93, Hewlett Packard Engineer Answer: 12.5 Solution: The triangle described consists of two right triangles: (20,21,29) and (15,20,25) joined on the 20 side. The point on the longest side (36) that is 20 from the opposite vertex is where the two triangles are joined. The shortest side (25) is part of the (15,20,25) triangle, and the distance from the right angle of a right triangle to the midpoint of the opposite side (hypotenuse) is simply half of the hypotenuse. (The hypotenuse can be viewed as one of the diagonals of a rectangle, and drawing a line from the right angle to the midpoint of the hypotenuse simply traces along the other diagonal of the rectangle, so they cross at the midpoint of each.) So the distance from the point to the midpoint of the shortest side (25) is 12.5.
Gregory Pack, Teacher Pensacola Catholic High School Pensacola, FL The distance is 12 1/2. Here is how I found it: Starting with an investigation of how far the opposite vertex was from the longest side, I constructed a perpendicular segment to the opposide vertex.By definition, this segment's length is the distance between the opposite vertex and the longest side. Let y equal this distance. This construction created two right triangles. Let x equal the leg of the right triangle on the left in my sketch. Therefore, 36 - x must be the length of the leg of the right triangle on the right in my sketch. Using the Pythagorean Theorem, I can create these two equations: 1) y^2 + x^2 = 25^2 which implies y^2 = 625 - x^2 and 2) y^2 + (36 - x)^2 = 29^2 which implies y^2 + 1296 - 72x + x^2 = 841 Substituting for y^2 in equation 2 using the y^2 of equation 1: 625 - x^2 + 1296 - 72x + x^2 = 841 combining like terms: 1921 - 72x = 841 subtracting 1921 from both sides: -72x = -1080 solving for x: x = 15 Substituting into equation 1): y^2 = 625 - 15^2 which implies y^2 = 400 Solving for y: y = 20 This tells us that the opposite vertex was exactly 20 units from the longest side. This implies there is only one point exactly 20 units from the longest side. Eureka! The vertex of the right angle of the two right triangles in my sketch is the point we are looking for. The shortest side of our original triangle is 25 units and is the hypotenuse for the right triangle on the left in my sketch. The midpoint of this side is 12 1/2 units from either end point. These end points also serve as vertices of our right triangle. The midpoint of the hypotenuse of a right triangle has an interesting property: it is equidistant from all the vertices. Therefore, the distance from our found point to the midpoint of the shortest side is 12 1/2.
Karen Petersen Teacher, Grades 9-12 School: Napa High School Answer: AB = 25, BC = 29, AC = 36, BD = 29, E is the midpoint of AB. At first, I wanted to determine where the altitude to AC was in relation to the line segment that had length 20. I called the altitude x and using Pythagorean Theorem, (AD)^2 + x^2 = 25^2 and (DC)^2 + x^2 = 29^2. It follows then that (DC)^2 - (AD)^2 = 29^2 - 25^2. Since AC = 36, then DC = 36 - AD. By substitution, (36 - AD)^2 - (AD)^2 = 216. Simplifying leaves 36^2 - 72(AD) = 216 and further 72(AD) = 1080. Thus, AD = 15. Since triangle ABD is a right triangle, AB = 25, and AD = 15, one finds that triangle ABD is a 3-4-5 triangle leaving x to be 20. It was surprising to find this to be the length of the altitude because this proves that BD is the altitude to AC! It is also important to note that there is only one line segment from B to AC that has length 20. Make the altitude from D to AB, call it F, assume it is located on AE. Using the process above or similar triangles, DF =12 and AF = 9. Since E is the midpoint, AE = 12.5, leaving EF = 3.5. This gives us the right triangle, EFD, with two sides known. Again applying the Pythagorean Theorem, ED = sqrt(12^2 - 3.5^2) = 11.48. One could also use repeatedly the Law of Cosines, but I try to approach problems the way most high school students would. I have never taught the Law of Cosines and I wonder if anyone has a good "proof" for high school students...
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