RIGHT ISOSCELES TRIANGLES ON THE GEOBOARD

RIGHT ISOSCELES TRIANGLES ON THE GEOBOARDAuthor(s): JOE DAN AUSTINSource: The Mathematics Teacher, Vol. 72, No. 1 (JANUARY 1979), pp. 24-27Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27961504 .

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RIGHT ISOSCELES TRIANGLES ON THE GEOBOARD

A wide range of problems flow from a simple isosceles triangle.

By JOE DAN AUSTIN

Emory University Atlanta, GA 30322

A rewarding mathematics experience for

many students is the discovery of mathe matical patterns in physical systems. Here we shall consider eight problems that arise when isosceles right triangles are con structed on a geoboard. The solutions can

be discovered by most students. However, it is interesting to see that the solutions can

be verified using only elementary algebra and geometry. The proofs yield some non

standard ways of showing several well known formulas.

The basic construction pattern involves

constructing isosceles right triangles on a

geoboard. Here all triangles will be isos celes right triangles. First construct such a

triangle with legs of length one. This

triangle will be called the basic triangle and also the (n

= l)st isosceles right triangle.

Construct this triangle in the lower left cor ner of the geoboard. Next add basic

triangles until an isosceles right triangle with legs of length 2 is constructed. (See fig. 1.) This is the (n

= 2)d isosceles right

triangle. Continue as shown until no larger triangle can be made on the geoboard.

The following eight problems refer to this series of isosceles right triangles. Stu dents are asked to complete tables for each

problem and then hypothesize a general pattern. Verification of the pattern is given for each problem.

Problem I: How many basic triangles make up the nth isosceles right triangle? (Refer to fig. 1 and table 1.)

General pattern: n2

Verification: There are n2 basic triangles. Consider the area of the main triangle, that

is, irt2. Since each basic triangle has area i there must be n2 basic triangles.

Problem 2: How many basic triangles must be added to the (n

- l)th isosceles

right triangle to produce the nth isosceles

right triangle? (See table 1 and table 2.)

Fig. 1. The first five nih isosceles right triangles

TABLE 1

Number of Basic Triangles in the nth

Right Isosceles Triangle

1 1 2 4 3 9

4 16 5 25

TABLE 2

Basic Triangles in: Number of -Basic Triangles

( - l)th Triangle nth Triangle Added

10 1 1 2 1 4 3 3 4 9 5 4 9 16 7 5 16 25 9

24 Mathematics Teacher



General pattern: 2n - 1 Verification: The general pattern sug

gested is 2/1 - 1. If we work again with the areas of the triangles, subtracting the area

of the (n -

l)th isosceles right triangle (i (n

- I)2) from the area of the nth isosceles

right triangle (i n2) gives i (2n -

1). Thus, 2n ? 1 basic triangles are needed. Note th^t the total number of basic triangles in the nth isosceles right triangle (n2) is the sum of the basic triangles added to go from the Oth

to the 1st, the 1st to the 2d, ? ?

?, (n -

l)th to

Arth isosceles right triangjes. This is simply

n2 = 1 + 3 + 5 + ? ? ? + (2n -

1).

Problem 3: A unit square is a square with sides of length one. If as many ur?it squares as possible are made in the nth isosceles

right triangle, how many basic triangles are not a subset of any unit square? (Refer to

fig. 2 and table 3.)

Fig. 2. Unit squares and the excluded basic triangles

TABLE 3

Number of Basic Triangles Not in Any Unit

Square

General pattern: Verification: The general answer can be

verified by noting that the hypotenuses of

the excluded basic triangles form the hy

potenuse of the main triangle. (See fig. 2.) The hypotenuse of the main triangle has

length /i\/2, and the hypotenuse of each

basic triangle has length y/2. Thus basic

triangles are not a subset of any unit

square.

Problem 4: In problem 3 how many unit

squares are there in the nth isosceles right triangle? (Refer to fig. 2 and table 4.)

TABLE 4

Number of Unit Square

0 1 3 6 10

General pattern:

Here the general pattern may not be so clear. Suppose we count the unit squares by rows. (See fig. 2 and table 5.)

TABLE 5

Sum of Unit Squares by Rows

0 1 1 + 2 1+2 + 3 1+2+3+4

0 1 3 6 10

General pattern:

Verification: In general, the solution is the sum of the first ? 1 positive integers. If the formula for this sum is known, we

have the general solution of ̂ ^

^n unit

squares. Again by using areas and prob lems 1 and 3, we see that a general proof is

possible. Subtract from the area of the

main triangle Q-/*2)

the area ?f the

basic triangles that are not a subset of any

unit square ^ ^.

This gives (n ~ ^n, or

the number of unit squares, since each unit

January 1979 25



square has area one. Note here that if we ask how many new unit squares are added in going from the (n

- l)th to the nth isos

celes right triangle and sum these up, we obtain the number of unit squares in the nth isosceles right triangle. This gives

^- ^ = 0+1 + 2+???+(?-1).

Problem 5: How many basic triangles have one or more sides that are a subset of a side of the nth isosceles right triangle? (See fig. 1 and table 6.)

TABLE 6

Number of Basic Triangles with a Side That Is a Subset of a Side of the nth Isosceles Right

Triangle

1 1 2 3 3 6 4 9

5 12

General pattern: 3{n -

1) if > 1 Verification: Note that basic triangles

make up each of the legs of the main

triangle and that by problem 3, basic

triangles make up the hypotenuse. Thus + + = 3n. However, if > 1, each basic

triangle that makes up a vertex of the main

triangle is a subset of two different sides.

Therefore, we have 3n - 3, or 3(/i ?

1), basic triangles if > 1. (Can you see why this argument fails for =

1?)

Problem 6: How many basic triangles have at least one vertex (but not a side) that lies on a side of the nth isosceles right triangle? (See figs. 1 and 3 and table 7.)

?1*6

Fig. 3. The (n =

6)th isosceles right triangle

26 Mathematics Teacher

TABLE 7

Number of Basic Triangles with a Vertex (but Not a Side) Intersecting a Side of

the nth Isosceles Right Triangle

1 0 2 1 3 3 4 6 5 9

General pattern: 3(n -

2) if > 2 Verification: This conjecture can be veri

fied in a manner similar to that used in

problem 5. Here there are ? 1 basic

triangles intersecting each side, and if >

2, three basic triangles intersect two sides.

(Again, do you see why > 2 is needed?)

Problem 7: How many basic triangles lie in the interior of the nth isosceles right triangle? (See fig. 1, fig. 3, and table 8.)

TABLE 8

Basic Triangles Entirely in the Interior

1 0 2 0 3 0 4 1 5 4 6 9

General pattern: (n ?

3)2 for > 2 Verification: Take the total number of

basic triangles, n2, and subtract the number of triangles that intersect the side of the nth isosceles right triangle, that is, the sum of the answers to problems 5 and 6. For > 2 we have

2 - (3(n

- 1) + 3(n

- 2))

= (n

- 3)2.

For = 1 and = 2 the principle applies but the formula fails. (Why?) Another way to verify the general pattern is to note that if the basic triangles intersecting the side of the main triangle are removed, an isosceles

right triangle with legs of length - 3

remains. Using problem 1, we find that there are (n

? 3)2 basic triangles in the

"interior" triangle.

Problem 8: This is the most difficult problem solved here. Using the right angle in the nth isosceles right triangle, make the



largest possible square that remains in the

triangle. Remove this square and observe that two isosceles right triangles remain. In each of these triangles repeat this process until no more squares can be made. (See fig. 4 and table 9.) (Note that the two

triangles may not be disjoint but that their intersection is at most a basic triangle.

Thus, no squares will contain this basic

triangle and lie in both triangles.) How

many squares result from this process?

Fig. 4. The three squares of the fourth isosceles

triangle and the seven squares of the fifth isosceles

right triangles. The first squares are separated from the

main triangles.

TABLE 9

Number of Squares

General pattern: Here it is doubtful that these few data suggest the general pattern. To obtain a general formula, let Sn be the number of squares that this process pro duces in the nth isosceles right triangle. Af ter the first square is removed, the two re

maining isosceles right triangles are

congruent and have legs of length

that is, the greatest integer func

tion of n ~*~ * . Each of these triangles then 2

contains Sr -, squares. Therefore, count

[ ] ing the square that was removed, we obtain a recursive relation

Sn = 1 + 2-S

[ ] for =

2, 3,

where Sx = 0. This relation permits us to

generate as much data as we need to see the

general pattern. The general solution for Sn is given as follows: For find the positive integer k so

then

< < 2*

Sn = 2* - 1.

You might try to prove this is indeed a

solution to the recursive relation. There are, of course, other problems we

could ask. For example, how many basic

triangles in the Aith right isosceles triangle have their right angle to the right of their

hypotenuse, as in figure 5? Perhaps you and

Fig. 5

your students can find similar problems and discover the resulting general patterns.

Want to encourage students to make discoveries with a calculator, get them to do more difficult and/or more realistic problems, and introduce them to new topics? Send your name and address for 2 free sample lessons, or send $20 to receive 80 worksheets which may be du plicated for classroom use. Suggested teaching tech niques and complete answer keys are included. Write:

lf+t0? 1229 Diana Lane *x Santa Barbara, CA 93103 'cu*

January 1979 27



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RIGHT ISOSCELES TRIANGLES ON THE GEOBOARD