Triangle ABC with area 20 - April 8-12, 1996
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Given: A(0,6), B(0,10), C(x,y), and the area of triangle ABC is 20.
- If triangle ABC is isosceles find C(x,y).
- Find all points C(x,y) such that the area of triangle ABC is 20.
Extra : Let A be (x,y) and B be (x1,y1). Now find the coordinates of C such that the area of ABC is 20. (This could be tough, as there will be some restrictions on A and B to make this work.)
Thanks to Carl Detzel of Shaler High School in Pittsburgh for suggesting most of this problem.
Annie says: Solutions
I really like this problem - except for the extra part. A number of people forgot that there is more to the coordinate plane than just the first quadrant! A couple of people even included the negative answer for the second part, but not the first...
There were a number of good explanations. The kids from Georgetown Day School did a really nice job in general. A couple of them (Josh and Nishant) made some nice inroads on the extra part as well. They explained pretty much what you'd want to do, but didn't take a stab at it (it gets complicated!). Thomas Kuo and Mike Sue also headed in the right direction.
To get an idea of how ugly the extra part could be, check out the solutions from Ilya Finkler and Ken Duisenberg. Here's an exercise to do in class: are their answers both right, and are they the same? (I guess they should be if they actually are right, but they look a little different - does it matter?)
Following are highlights. This week we welcome new contributor Ilya Finkler. The names of all the people who submitted correct solutions and most of the solutions are also available.
Josh Hersh Grade 8 Georgetown Day School, Washington DC Teacher: Paul Nass 1. If triangle ABC is isosceles find C(x,y). First I located points A and B and connected them. Then I found the median of line AB which was (0,8). Since the triangle had to have an area of 20, the base (4) times the height (x) divided by 2 equals 20. 4x/2=20 2x=20 x=10 I found the point ten spaces to the right of (0, 8) and also ten spaces to the left of (0,8) and found that point C= (10, 8) or (-10, 8) 2. Find all points C(x,y) so the area of triangle ABC is 20. I redrew points A and B and connected them. I redrew the isosceles triangle and also drew some other triangles with a height of ten - for example 2 right triangles one with A as the right angle and the other with B as the right angle. By playing around with the location with C for a while keeping the height at 10, I discovered that the coordinates of C were (10, y) and (-10, y). Extra. I wasn't able to really prove anything mathematically, but by putting A and B in different places and using the height of ten and the distance formula I found that point C could be any point along a line running through a point that is the distance of the height away from line AB and has the same slope, or is parallel to line AB.
Nishant Kumar Grade 8 Georgetown Day School, Washington DC Teacher: Paul Nass 1. C(10,8) or -10,8) 2. In problem 2, C can be anything on the line with x coordinate 10, the y coordinate is anything along this line or the line along the x coordinate of -10 Extra: Given any point a and point B, you derive the slope of the line using the slope formula(y2-y1)/(x2-x1), since you intend to draw a line parallel to the line between A and B you will make a line with the same slope, now all you need to do is find out the distance between the lines which is also determined by the length of line B, the formula for this is 40/ the square root of (x2-x1) + (y2-y1)
Thomas S. Kuo Grade: 7 School: Murray Junior High School, Ridgecrest, California 1. C(-10,8) and C(10,8) (1) The distance from (0,6) to (0,10) was 4 and its middle point was (0,8). (2) Since area = 20 = (1/2) * d * h, then h = 10. (3) The point C which has a distance 10 from (0,8) and forms an isosceles triangle will be (10,8) and (-10,8). 2. Any point along the lines x= 10, and x=10 is the solution, because those points can make a height 10 to segment AB. Extra. Let d be the distance between A(x,y) and B(x1,y1), then d = sqrt((x-x1)^2+(y-y1)^2) d should be 0 < d < 20 to form a triangle with area 20. Again, area = 20 = (1/2) * d * h, and h = 40 / d. The point C should be on lines which parallel to AB and keeps a distance h from AB. There are two lines that meet these requirements. One is on each side of AB.
Mike Sue Grade: 10 School: Granada 1. For an isosceles triangle, line segment AC and segment BC must be equal in length. y=6+10/2=8 A=1/2bh 20=1/2(10-6)|x| 40=4|x| |x|=10 x= +/-10 2. Find all points C(x,y) so the area of triangle ABC is 20. A=1/2bh 20=1/2(10-6)|x| 40=4|x| |x|=10 x= +/-10 Extra: Let A be (x,y) and B be (x.,y.). Find the coordinate of C such that the area of triangle ABC is 20. If y=y., then A=1/2bh 20=1/2|x,-x|h h=40/|x,-x| C(second x, y+40/|x,-x| or C(second x, y-40/|x,-x| WHERE SECOND X IS ANY REAL NUMBER
From: Ilya G. Finkler Grade: 12 School: Liberty High School, Youngstown, OH 1. A=a*h/2=>A=AB*x/2 AB=4 and A=20=>x=2*20/4=10 [and -10 as well] B(0,10) a) BC=AC. Angle CDB is right, then l BD=DA. Then point C has coordinates l l (10,(10+6)/2)=(10,8)[and (-10,8)] l l l l b) Since the height (CD) is equal to l l x-coordinate of point C, then point l l C has coordinates (10,y) and (-10,y), D(0,y) ll l l l l l C(x,y) where y is any real number. l l c) A(x,y) and B(x1,y1)=>line AB has l l l l equation f(z)=(y1-y)/(x1-x)*z+y1-(y1-y)/(x1-x)*x= l l =(y1-y)/(x1-x)*z-(y1*x+y*x1)/(x1-x) l l From the previous parts of the problem we can l conclude that point C lies on either of two lines A(0,6) that are parallel to AB. We can also conclude that the distance between either of these lines and AB is 2*20/SQRT((y1-y)^2+(x1-x)^2). Let's call one of these lines CE. l Y \ CD=40/SQRT((y1-y)^2+(x1-x)^2) l \ CN is parallel to OY(y-axis) l \ C Angle CDB is right. l\ /\ Tan(angle ALX)=(y1-y)/(x1-x) l \ /''\ [if x=x1 GOTO b)] l \ / '' \ CN=CD/cos(angle DCN) l A \ / '' \ If we continue line CD till it crosses x-axis l \ / '' \ at point Z, then angle CZX will be equal l \ / '' \ E angle ALX-Pi/2 [if angle ALX is obtuse] or Pi/2+ l D \ '' angle ALX [if angle ALX is acute]. l \ '' Then angle ZCN(=angle DCN)=Pi-angle ALX if angle ALX l \ '' is obtuseor angle ZCN=angle ALX if angle ALX is acute. l \ '' Cos DCN=-Cos ALX (ALX-obtuse) or Cos DCN=Cos ALX l \ '' (ALX-acute). We can conclude Cos DCN=ABS(cos ALX) l B \'' ABS(cos ALX)=1/ABS(sec ALX)=1/SQRT(1+(tan ALX)^2)= l \' 1/SQRT(1+((y1-y)/(x1-x))^2)=ABS(x1-x)/SQRT((y1-y)^2+ l N\ (x1-x)^2)CN=(40/SQRT((y1-y)^2+(x1-x)^2))/(ABS(x1-x)/ l \ SQRT((y1-y)^2+(x1-x)^2))==40/ABS(x1-x) l \ l----------------\-------> O L\ X Then C lies on either of lines: f(z)=(y1-y)/(x1-x)*z-(y1*x+y*x1)/(x1-x)+(or "-")40/(x1-x) [we can drop ABS because of "+"/"-"]
Ken Duisenberg Post-Master's School: Stanford; work at Hewlett Packard Answers: 1. (10,8) or (-10,8) 2. x=10 or x=-10, y = anything EXTRA: C(x3,y3): (y3-y1)*(x1-x0) = (x3-x1)*(y1-y0) +/- 40 Solution: Triangle area = bh/2, for any base b of the triangle and height h is measured perpendicular to the base. So let base b = AB = 4. Then 20 = 4h/2 --> h=10. For (1) and (2), the last point must be 10 away from base b (x=10 or -10). For (1), y=8, so C = (10,8), or (-10,8). For (2), y = any value, so C = (10,y) or (-10,y) EXTRA: I reassigned A as (x0,y0), and tried to find C(x,y). Similar to (2), we need to find a line parallel to AB at a distance h that will result in bh/2 = 20, where b = distance between A and B. Obviously, there will be two lines, one on either side of AB. The only restriction on A and B is that they are not the same point. To solve, if we can find BD perpendicular to AB, with length of BD = h, then we can pass a line parallel to AB through D, and that line will be the line we're looking for. C could be anywhere on the line. First, b = sqrt[ (x1-x0)^2 + (y1-y0)^2 ], and Area = bh/2 = 20, so h = 40/b. Next, pick a point D(x2,y2) to satisfy the following: (x2-x1) = (+/-)(h/b)*(y1-y0), OR x2 = x1 +/- (40/b^2)*(y1-y0) (y2-y1) = (-/+)(h/b)*(x1-x0), OR y2 = y1 -/+ (40/b^2)*(x1-x0) A quick analysis shows: (y2-y1)/(x2-x1) = -(x1-x0)/(y1-y0) -> BD is perpendicular to AB. (y2-y1)^2 + (x2-x1)^2 = h^2 -> length of BD is h, as desired. Now define a line parallel to AB through D: (y-y2) = [(y1-y0)/(x1-x0)]*(x-x2) Substituting: (y - y1 +/- (40/b^2)*(x1-x0) ) = [(y1-y0)/(x1-x0)]*(x - x1 -/+ (40/b^2)*(y1-y0)) Or, in y=mx+b form: y = [(y1-y0)/(x1-x0)]*x + y1 - [(y1-y0)/(x1-x0)]*x1 -/+ (40/b^2)*[(y1-y0)/(x1-x0)](y1-y0) -/+ (40/b^2)*(x1-x0) y= [(y1-y0)/(x1-x0)]*x + y1 - [(y1-y0)/(x1-x0)]*x1 -/+ (40/b^2) * (b^2 / (x1-x0)) FINALLY: y= [(y1-y0)/(x1-x0)]*(x-x1) + y1 -/+ 40/(x1-x0) This can be rewritten as: (y-y1)*(x1-x0) = (x-x1)*(y1-y0) -/+ 40 Any C(x,y) satisfying the above equation will provide the desired area of the triangle. Notice that "40" can be replaced by "2*Area" for any other area. The multiplication by (x1-x0) above is not strictly legitimate if x1=x0, but by trying it out on the original problem, we find the result is valid: In (2) above, x1=x0=0, so the equation becomes 0 = 4x -/+ 40, or x = 10 or -10.
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