Geometry Forum - Problem of the Week
Solutions - Constructing an Isosceles Triangle, Feb. 27-March 3, 1995
Annie says:This week's problem provided a glimpse into a broad spectrum of ideas about constructions! Three examples of different approaches: Ruth Carver's students submitted wonderful, "classic" constructions of isosceles triangles. Jim Swift's students from Burnaby South Secondary submitted some classic constructions, but then also went wild with "alternative" constructions - check out their answers. Several solutions from another school indicated that they hadn't done constructions at all - the students I talked to figured that you could construct infinitely many isosceles triangles by making sure two sides were congruent, but didn't really understand when I asked how many different ways could they do that. Very interesting to see the different reactions!
As alluded to above, a few solutions exhibited some confusion between being able to construct infinitely many things (isosceles triangles in this case), and being able to construct something infinitely many ways. Sure, you can construct infinitely many isosceles triangles by connecting the center of a circle with two points on the circle, but that's only one construction! At first, there seemed to be a line between those who got it and those who didn't, which was dependent on class, but then I received both sorts of solutions from students in one class. Hmmmm...
And here are the solutions. Ruth's kids sketches are available on our Web site on the K12 page - do check them out if you have access to that resource. The sketches are also available for downloading in our archives. The path is ftp://mathforum.org/problem.of.the.week/solutions/sketches/mar.3.95/.
Experimental Math Class
This week's problem was given as a collaborative problem to our class. This is the combined efforts from the Experimental Math Class at Burnaby South SS.
1. Construct a segment AB, Draw a bisector line CD. Any points on CD
linked to AB make isosceles triangle.
2. A geometry set triangle with two 45 degree angles and one right angle.
3. Take 2 points on a circle and connect the two and the center. MV
4. A square and draw a diagonal.
5. Trace the face of a 4-sided die.
6. Trace a clothes hanger if you carry on the line MV.
7. Draw the diagonal of the rhombus SD.
8. An elastic band pulled by a force in its middle.
9. Use a GameBoy button as a stamp and join two corners.
10. When an umbrella opens up, the spines form an isosceles triangle.
11. Use graph paper to draw any isosceles right triangles.
12. Construct two parallel lines (or one). Use a compass at a point A to
draw a radius which is long enough to pass the other parallel line on BC. Link AB and BC.
13. By placing two piece of paper edge to edge and draw a line at the
other edge. (M4)
14. Take two sticks of equal lengths, join them at one end. Add third
side.
15. Draw 3 equal circles tangent to each other. Join 3 centers. 16. Take a 6-sided star and you have 6 isosceles triangles.
Here at Burnaby South Secondary School, the Problem of the Week for February 27 - March 3 was done as a collaborative group effort. Our solutions are being sent in two batches, of which this is the second (or should be):
1. Given a horizontal rod. To this rod, tie the midpoint of a string, to either end of which is attached a weight of equal mass. Let the weights drop at the same time, and take a photograph. Trace over the photograph:
______________________________
-------------*----------------
| / \ |
| / \ |
\ / \ /
-> / \ <-
/ \
/ \
Weight 1 * * Weight 2
2. Given 2 points A and B, construct a point X such that AX is
perpendicular to BX. Reflect B over AX:
|
* A
/ |
/ |
/ |
/ |
B/ |X B'
-*------*------*-
|
3. Given any right angle, make a 45 degree angle to the original
corner:
|
|\
| \
| \
| \
| \
| \
|_ _\
GIVEN->| | |45\ <-Construct with protractor, etc...
----------
4. Given a line AB, rotate the original line around point A so you
have a line AB'. Triangle is ABB':
---
-*- B'
---
---
A --- B
-*------------------*-----
5. Attach a weight to a string and measure out equal distances along
the string at either end of the weight. On a bulletin board, stick in 2 drawing pins and over each hang the string at one of the pre- measured points. Trace:
____________________________
| |
| |
| __ __ |
| *\- - - - -/* |
| Pin 1 \ / Pin 2 |
| \ / |
| \ / |
| \ / |
| * Weight |
|____________________________|
6. Given a parabola, find the 2 points where the graph crosses a
certain horizontal line (i.e: y=0), and the vertex. Connect the three points. (Make sure the vertex is not on the line you have chosen)
7. Put a cone outside on a sunny day. Trace its shadow.
8. Given a non-rectangular parallelogram, trace one of its non- horizontal sides. Flip the parallelogram over a vertical line, and re-trace that side. Extend the lines until they meet:
\|/
/|\
/ | \
______/ | \______
/ / | \ \
/_____/- - |- - \_____\
/ | \
9. Find an isosceles triangle in a book and trace it.
10. Fold a piece of paper exactly in half, cut along the diagonal of the half-piece, and open the paper out.
11. Draw 2 perpendicular lines, and a square around them. This forms 8 isosceles triangles.
12. Draw a perfect 5-pointed star, with a pentagon in the middle. This forms 5 isosceles triangles.
13. Draw a pentagon around the star. You now have 10 more isosceles triangles.
As well, it was suggested that many things that you see outside are already in the form of isosceles triangles. They figured we could simply take a photograph of these objects, and trace the photographs onto paper. Some of the ideas were:
-Teepee
-Roof of a doghouse
-Pine tree
-Funnel
[Sean was prompted for some explanations of a couple of their answers]
Hi Annie!
Sorry - you're not the only one late in replying to e-mail. I got your letter of the 20th just now - off for spring break last week - and, in reply to your concerns:
1. Okay. The definition of a circle is the set of all points equidistant from a given point (the center, with the distance between the center and all the other points being the radius). Therefore, for any point on the circle (as the endpoints of the chord are), they will all be equidistant from the center.
2. There is a theorem I learned in school a few years ago stating that all points equidistant from the endpoints of a given segment lie on the perpendicular bisector of the segment, and vice versa. It runs more or less as follows:
*A
|
*N
|
D |B E
*--------------------*--------------------*
|
|
|
|
*C
Given that any point N is equidistant from D and E, triangle DNE is
isosceles (DN=NE), therefore, angle NDE=angle NED. NB, of course, is equal to NB (reflexive property of congruence), and so triangle DNB is congruent to triangle ENB (ASA theorem - I hope I don't have to explain that one). DB=BE (1) (corresponding parts of congruent triangles are congruent), and angle NBD=angle NBE (same). Since DE is a straight line, NBD+NBE=180 degrees, and they are both right angles (2).
So, line NB is the perpendicular bisector of DE.
Sorry I didn't put this in my original response, wasn't thinking, I guess. Thanks,
- Sean Nichols.
Uyen Nguyen
Solution to POW for Feb.27-March 3Submitted by Uyen Nguyen, Steel Valley High School, Pa.
Method 1:
1. Draw a segment of any length.
2. Open a compass to any setting greater than 1/2 of the length of the original
segment and lock it on that setting.
3. Put the compass point on one of the endpoints of the segment and mark an arc
above the segment.
4. Repeat putting the compass point on the other endpoint. 5. Connect the endpoints to the intersection of the arcs.
Method 2:
Construct a square or rhombus and bisect it by connecting two opposite vertices. This will make two congruent isosceles triangles.
Method 3: (Construct two congruent right triangles that share a leg.)
1. Use a protractor to construct a 90 degree angle.
2. Connect the endpoints of the two segments that formed the angle.
3. Measure AB with a compass and keep it on that setting.
4. Put the compass point on B and mark an arc to the right.
5. Measure AC with a compass and keep it on that setting.
6. Put compass point on C and mark an arc that would intersect the arc to the right of B.
7. Connect C and B to new intersection.
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