I like this! Thanks for sharing. :) Do you make kids find perpendicular bisectors via folding vertices on top of each other, or via rulers and protractors/compasses?And, have you seen the NCTM illuminations lesson on pizzeria borders? http://illuminations.nctm.org/LessonDetail.aspx?id=L745 That one works well as a multi-day exploration of circumcenters and perpendicular bisectors.
Love it! And I'm totally stealing it for my classes. :) Thanks!
I know you said on Twitter that you didn't want to brag about this lesson, but I really think you should.
Hummmmm ... I work in a multi-story building - I wonder if they can deal with 3D? Wanna try.
I want to assume you let them run outside to get the treasure, because I'm pretty sure that's your style, but that would then raise some other logistical questions for me.
@Mimi I had not seen that activity - I like how it generalizes the idea to more than three points. Thanks for sharing.@Curmudgeon You going to put a treasure between floors? Exciting. Maybe in a stairwell.@Dan I didn't let them out during class. I used the same M.O. as the locus star search from last year. The "treasure" was a cardboard star taped to a wall in a hallway. They could go look for it after class and upload a picture.
@Kate: the points equidistant from two points lie along a line (perp bisector and all that) and thus the third point in 3D introduces 2 suitable points equidistant. (I think -- must make sure of this) So, if I play my points right, I could get locus on the second or third floor and the other in the basement.
The sheer genius of this lesson must not, for the good of the republic, be understated.I had not seen the locus star search post previously, but I am definitely stealing it. With props to La Nowak, of course.
Nitpick, curmudgeon: if you're going to move to 3D, the pointe equidistant from two fixed points form a plane, not a line. Adding a third point leads to the intersection of three planes.
@Curmudgeon:In 3-space, 3 points (A, B, C) will define a plane. The circumcenter for triangle ABC will be the point (O) equidistant from A, B, and C in that plane. However, you can travel along the line through O and normal to the plane and all points along that line will be equidistant from A, B, and C.
This is a nice application of the circumcenter.Looking at the map of the school, it makes me think of taxi-cab geometry. If you looked at actually "hallway" distances from these three points, where would the treasure be?
I love it. I solved it by loading the image into GeoGebra and constructing the circle through three points. Image
@hao, @mathochistYeah, I noticed that when I was trying to setup the numbers so that pythagorean triples would apply. Unfortunately, I got mentally wrapped up in the intersection of spheres instead ( I know, backwards)Anyway.Three points on a floorplan are coplanar (duh) and so the locus of points equidistant lie along the line perpendicular to the plane that passes through the intersection of the p-bisectors of AB, BC, and AC. So, if I go 3d, I need a fourth point. Must think.
Slick idea -- I'd hug you but know about your feelings on the matter.About 400 ninth graders at my school will participate in a course-wide treasure hunt next Friday. My kids will take a clue (consisting of 3 corners) and construct all 4 triangle centers. At which point they unlock the second half of the clue that eliminates 3 of the centers and points them to a treasure's general location. I'll post a complete debrief after the event. Until then, here's the GSP file I'm working in: http://dl.dropbox.com/u/2142796/Clarkston%20Upstairs%20Floorplan.gsp--Megan Golding
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