on inscribed and escribed circles of right triangles, circumscribed triangles, and the four-square,...

On Inscribed and Escribed Circles of Right Triangles, Circumscribed Triangles, and the Four-Square, Three-Square ProblemAuthor(s): David W. HansenSource: The Mathematics Teacher, Vol. 96, No. 5 (MAY 2003), pp. 358-364Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/20871339 .

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David W. Hansen

On Inscribed and Escribed Circles of

Right Triangles, Circumscribed Triangles, and the Four-Square, Three-Square Problem

or more than twenty years, I have been studying the

fascinating relationships between right triangles and their inscribed and escribed circles. An inscribed circle is one that is tangent to all three sides of a triangle and whose center lies inside the triangle. An escribed circle is one that is tangent to one of the sides of the triangle and to the extensions of the other two sides and whose center lies outside the

triangle. Every triangle has one inscribed and three escribed circles, as shown in figure 1.

This article discusses right triangles only. Let triangle ABC be a right triangle with legs a and 6 and hypotenuse c. Let r be the radius of the inscribed

circle; and let ra, rb, and rc be the radii of the escribed

Fig. 1 Inscribed and escribed circles of triangle ABC

circles tangent to sides a, b, and c, respectively. Fig ure 1 shows a typical right triangle with its inscribed and escribed circles and associated radii.

At this point, the reader can take pencil and

graph paper and construct a 3-4-5 right triangle with its inscribed and escribed circles. Measure their radii to see whether you can discover any interesting relationships between the radii and the sides. I guarantee that you will find some. In draw

ing these circles, note that their centers all lie on the two lines that pass through vertex C and that make 45 degree angles with side a extended.

A teacher can split the class into groups of three students or more and give each group a different

right triangle to construct. For example, one group constructs a 3-4-5 right triangle; a second group, the 5-12-13 right triangle; a third group, the 7-24-25 right triangle, and so on. Students can easily draw the inscribed and escribed circles using graph paper, pencil, compass, and straightedge by simply estimat

ing the location of the centers of the circles by trial and error. If the teacher prefers and if the class has

enough time, this activity presents an opportunity to use classical straightedge-and-compass construc tions. Alternatively, geometry software such as The Geometer's Sketchpad or Cabri Geometry II can be

used; the teacher asks students to draw the triangle with its sides on the axes and hypotenuse in the second quadrant. Then, using circles with centers

lying on the lines y = and y = - , the students can

try various values for the radii until all circles are

tangent to their respective lines.

Next, ask each group to measure the length of the radii of the inscribed and each escribed circle and to record their results (and verify them with

David Hansen, [email protected], has taught for thirty three years at Monterey Peninsula College in Monterey, CA 93940. He is interested in algebraic number theory and delights in using lower-division college mathematics to obtain surprising results.

358 MATHEMATICS TEACHER



the teacher) in a table similar to table 1 before they try to discover any patterns or relationships among the various radii and sides.

TABLE 1

3 4 5 5 12 13 7 24 25

2 3 3 10 4 21

6 15 28

Some simple discoveries might be

or

ra + rb = c.

The teacher might then ask each group the fol

lowing questions:

What relationships exist between just the inscribed and escribed radii themselves?

What relationships exist among just the sides?

What relationship exists between the radii and the sides?

What have you discovered about the reciprocals of the radii?

What have you discovered about the squares of the sides and radii?

What are the relationships between the circum ferences or areas of the inscribed and escribed circles?

Connect the centers of the escribed circles to form a triangle. What special properties, if any, does this triangle have?

After the students have discovered some of the

relationships between the sides and radii, the teacher might ask them to justify their conjectures by applying their ideas to such other right triangles as the 8-15-17 or even the 9-40-41 or 12-35-37 tri

angles. When students see that their ideas apply to these right triangles also, they might wish to prove that they apply to all right triangles. By using the formulas given subsequently, they should be able to

verify many or all of their conjectures and experi ence the pleasure that I?and I hope you also?get from studying these fascinating inscribed and escribed circles of right triangles.

My interest in escribed and inscribed circles of

right triangles goes back to 1979, when in Hansen

(1979), I showed that

(1)

(2)

(3)

r = - a + b-c

-2

a-b +c

b-a + c

and

(4) a + b + c

By adding equations (1), (2), and (3), I proved that

(5) r + ra + rb =

rc,

giving the following theorem.

THEOREM A. The sum of the radius of the inscribed circle and the radii of the two escribed circles tan

gent to the legs of a right triangle equals the radius of the escribed circle tangent to the

hypotenuse of the right triangle.

I also discovered that

(6) rrc=rarb _ ab '

2 = area of triangle ABC,

which can be stated as theorem B.

THEOREM B. The product of the radii of the inscribed circle and the escribed circle on the hypotenuse of a right triangle equals the product of the radii of the escribed circles on the two legs, and this prod uct equals the area of the right triangle.

The proofs of these two theorems involve only algebra and geometry and can be readily used by teachers to illustrate that simple applications of these subjects can lead to interesting and satisfying results. Using the areas of appropriate triangles (see Hansen [1979]) constructed in each of the escribed circles and inscribed circle, together with the Pythagorean theorem and the quadratic formu

la, yield the desired equations. I have been fortunate in discovering many more

wonderful theorems about these fascinating circles. The proofs of these theorems are all accessible to students studying algebra, geometry, and analytic geometry. Some of these theorems follow.

THREE REMARKABLE THEOREMS When I examined equations (1), (2), (3), and (4), I was struck by their symmetry and simplicity. Per

haps other theorems might be discovered by exam

ining them more closely. Adding equations (1), (2), (3), and (4) together

leads to the following result:

r + ra + rb + rc a + 6-c a-fc+c b-a+c a + ? + c

2 2 2 2 a+b-c+a-b+c+b-a+c+a+b+c

2

2a + 26 + 2c

The proofs of these theorems are

accessible to students

taking algebra, geometry, and analytic geometry

Vol. 96, No. 5? May 2003 359



In a right triangle, the

sum of the inscribed

radius and

the three escribed

radii equals the perimeter

of the

triangle

_2(a + fc + c) 2

= a + 6 + c.

Therefore,

(7) r+ ra + rb + rc = a + b + c.

The first remarkable and satisfying result is that the sum of the radii equals the sum of the sides. Theorem 1 reads as follows:

THEOREM 1. In a right triangle, the sum of the inscribed radius and the three escribed radii

equals the perimeter of the triangle.

Examining the squares of equations (1) through (4) leads to the following:

2_(a + ft-c)2 4

_q2 + fr2 + c2 + 2a6-2ac-2?c 4

? (a-? + c)2

_ a2 + b2 + c2 - lab + 2ac - 2bc 4

?2_(&-a + c)2 ?-4

a2 + ?2 + c2-2a?-2ac + 2?c.

(a + b + c)2

a2 + b2 + c2 + 2ab + 2ac + 2?c

Adding all four equations together gives

r2,r2,r2,r2_ 4a2 + 4?2 + 4c2 + 0 + 0 + 0

? + ? + ? = a + r? + i adding the first two reciprocals ra rb rc rarb rc

_ ra + rb + r multiplying the numerator and denominator of

rar? 7TC the third reciprocal by r

r<l + rb +~r using theorem to replace rarb with rrc

adding the fractions and rearranging terms

rrc rr,

J + ra + rh

rrr

.1 r

using theorem A to replace r + ra + r6 with rc

dividing the numerator and denominator by rc

Fig. 2 Proof of theorem 3

or

(8) r2 + r2 + r2 + r2 = a2 + ?2 + c2t

The second, and definitely remarkable, result is stated as theorem 2.

THEOREM 2. Ina right triangle, the sum of the

squares of the inscribed radius and the three escribed radii equals the sum of the squares of the

lengths of the sides of the triangle.

In any right triangle,

r + ra + rb + rc - a + b + c.

Even more remarkably,

r2+ rl + r2 + r2 = a2 + b2 + c2.

Finally, readers can examine the reciprocals of the inscribed and escribed radii. By theorem A, r + ra + r? = rc. However, since r is the smallest of the four radii and since rc is the largest, 1/r is the

largest of the reciprocals and l/rc is the smallest; and readers thus cannot expect that

1 1. 1_1 r ra rb~rc'

But if 1/r is interchanged with llrc in the preceding equation,

1111 - +- + - = -, ra rc r

the result might be valid. Figure 2 shows the proof of this conjecture.

As figure 2 indicates, the conjecture is true, proving the remarkable result shown in theorem 3.

THEOREM 3. In a right triangle, the sum of the recip rocals of the three escribed radii equals the recip rocal of the inscribed radius.

Truly, studying the radii of the inscribed and escribed circles of a right triangle has revealed some beautiful theorems, and only a knowledge of

beginning algebra was needed.

Summarizing, Theorem 1:

r + ra + rb + rc = a + b+ c

Theorem 2:

r2 + r2 + r2 + r2 = a2 + b2 + c2

Theorem 3:

1111 ? + ? + ? = -.

ra rb rc r

The following examples apply theorems 1, 2, and 3

to the 3-4-5 right triangle.




Example 1

Let a = 3, b = 4, and c = 5. Then

r 3+4-5

= 1,

r? =

fr

anti

3- 4 + 5 2

2,

4- 3 + 5 2

3,

3 + 4 + 5

So

and

r + ra + rb + rc =1 + 2 + 3 + 6

= 12,

<z + 6 + c = 3 + 4 + 5

= 12,

which verifies theorem 1.

Also,

r2 + r2 + r2 + r2 = 12 + 22 + 32 + 62 = 50,


Finally,

_6 "6

= 1,

and

1_I r~l

= 1,


Example 2

This example attempts to derive a relationship be tween the circumferences and areas of the inscribed and escribed circles, if possible. Multiplying both sides of equation (5) by 2n gives

2/rr + 2wra + 2jxrb -

2wrc>

or the sum of the circumference of the inscribed cir

cle and the circumferences of the two smaller escribed circles equals the circumference of the

largest escribed circle.

Multiplying both sides of equation (8) by /r gives

nr2 + nr\ + nr\ + nr\ = (a2 + b2 + c2), = ;r(c2 + c2), (since a2 + b2

- c2)

= 2;rc2,

or the sum of the areas of the inscribed and escribed

circles equals 2n times the square of the hypotenuse.

INTERESTING RELATIONSHIPS

Many interesting relationships between the

inscribed and escribed radii and the sides of a right

triangle may be discovered. The following results can all be proved by using equations (1), (2), (3), and (4) and simple algebra.

1.

3.

r + ra = a

r + rb = b r + rc

= a + b

ra + rb = c

ra + rc = a + c

rb+ rc = b + c

ra + b = rc or

a + rb -

rc or rc-b

= ra,

rc-a =

rb, (9) r + c = rr or r,-c = r.

THE AREA OF RIGHT TRIANGLE ABC Let be the area of any right triangle ABC. From equation (6), rrc = and rarb = K. Multiplying these two equations together gives rrcrarb = K2, or K2 =

rrarbrc. Thus, - V/rar6rc, a very nice result that

gives theorem 4.

THEOREM 4. 77ie area of a right triangle is the

square root of the product of its inscribed radius

and its three escribed radii.

A different (and perhaps more interesting) way to

obtain theorem 4 is to use Heron's formula, which

states that the area of any triangle is given by = Ms - a)(s - b)(s

- c), where s = (a + b + c)/2, the

semiperimeter of triangle ABC. Since in any right

triangle, rc = (a + b + c)/2, then rc = s. Substituting

in Heron's formula gives the following result:

(10) = Vrc(r -

a)(rc -

?)(rc -

c)

As previously indicated for equations (9), rc - a =

rb,rc-b =

ra, andrc-c = r.

Substituting these equations into equation (10)

gives

K=^?rcrbrar

= ^rrarbrc,

and theorem 3 has been proved.

THE CIRCUMSCRIBED TRIANGLE Let Ea, Eb, and Ec be the centers of the three

The area of a

right triangle is the square root of the

product of its inscribed radius and

its three escribed radii

Vol. 96, No. 5 ? May 2003 361



Fig. 3

Triangle formed by connecting the three centers of the tangent circles to triangle ABC

escribed circles tangent to sides a, b, and c, respec tively, of right triangle ABC. Consider the triangle formed by connecting these three centers with

straight lines, as shown in figure 3? This triangle, EaEbEc, appears to pass through (or contain) the three vertices of right triangle ABC.

First, place right triangle ABC on an x-y coordi nate system so that C is at the origin, A is on the

positive y-axis at (0, 6), and is on the negative x-axis at (-a, 0).

Next, notice that Ea is at (-r0, -rj, Eb is at

(r&, rb\ and Ec is at (-rc, rc), since each of the escribed circles is tangent to the x- and y-axes.

The slope of

_ rh + ra EaEb

and the slope of

CEh

= 1,

rb-0

1.

The slopes of EaEb and CEb are equal, and EaEb and CEb have a point in common. Therefore, Ea) C, and Eh are collinear, so C lies on EaEb and on tri

angle EaEbEc.

The slope of

(11) EbEe = ?L

-a + c

by using the appropriate equations. The slope of

b - (a + ? + c) 2

a +b +c 2

Multiplying the numerator and denominator by 2

gives 2b-a-b-c

a + b + c so that

(12) AEC: -a + b-c

a + + c

Showing that equation (11) equals equation (12) will prove that Eb, A, and Ec are collinear, and thus A lies on line EbEc. See figure 4,

Thus, equation (11) equals equation (12), and A is a point on line EbEc. In a like manner, proving that ? is a point on line EaEc is possible. Theorem 5 is therefore true.

THEOREM 5. The triangle EaEhEc passes through, or

contains, the vertices A, B, and C of the right tri

angle ABC.

We call triangle EaEbEc the circumscribed tri

angle associated with right triangle ABC, since it contains vertices A, B, and C of right triangle ABC and since all other points of right triangle ABC lie inside triangle EaEbEc. What about the lengths of the sides of the cir

cumscribed triangle? The distance d(Ea, Eb) from

Ea to Eb is given by

(13)

d(Ea,Eb) = ^raf + (rb + raf

= V2(r + r6)2 = (ra + r6)V2 = cV2.

The previous step used the formula ra + rb = c

from result 2 under the "Interesting Relationships"

heading. Thus, the length of side EaEb equals the length of

the hypotenuse of triangle ABC multiplied by V2. Using the distance formula and equations (2),

(3), and (4), the lengths of the other two sides are

d{Ea,Ec) = mUrJ) = V(a + c)2 + b*

MATHEMATICS TEACHER



and

d(Eb,Ec) = i2(rt + r2) = Va2 + (? + c)2.

AREA OF THE CIRCUMSCRIBED TRIANGLE A nice formula for the area of the circumscribed tri

angle can be obtained by noting that the slope of

EaEb = 1, as shown previously, and that the slope of

the line segment from Ec to the origin O is

Thus, EcO is perpendicular to EaEb and hence is

perpendicular to the base EaEb and is the altitude

of triangle EaEbE .

The length of altitude EcO is equal to

V(-rc -

O)2 + (rc - 0)2 = V2r?

= rcV2,

and the length of base EaEb = cV2, from equation

(13). Then the area of the triangle is

^J(base)(altitude) = ^)(cV2)(rcV2) = crc,

and theorem 6 has been demonstrated.

THEOREM 6. The area of the circumscribed triangle

of right triangle ABC equals the product of the lengths of the hypotenuse and the radius of the

escribed circle tangent to the hypotenuse, that is, K =

crc.

NUMBER-THEORETIC CONSIDERATIONS The four-square, three-square problem

Can seven positive integers be found such that?

q) the sum of the first three integers equals the

fourth, and

b) the sum of the squares of the first four integers

equals the sum of the squares of the last three

integers?

How about the integers 1, 2, 3, 6, 3, 4, 5?

Clearly, part (a) is satisfied, since 1 + 2 + 3 = 6.

Also, part (6) is satisfied, since

l2 + 22 + 32 + 62 = 32 + 42 + 52.

Thus, it is a solution of the four-square, three

square problem. But how would you solve the general problem of

finding seven integers a, b, c, d, e, f, and g, such that

q+?+c=d

from equation 11

multiplying the numerator and the denomi

nator by a + b + c

multiplying out the numerator

substituting b2 - c2 for hi2, since a2 + b2 - c2, and factoring the rest of the numerator

factoring the difference of squares and

commuting the second term

factoring out b + c and rearranging

canceling + c from the numerator and

denominator

Fig. 4

Proof that equation (11) equals equation (12)

and

a2 + &2 + c2 + ?2 = e2+/-2+g2?

When one knows the relationships between the

inscribed and escribed radii and the sides of a right triangle, the solution is straightforward. Simply

pick any right triangle with integral sides. Then use formulas (1), (2), (3), and (4) to find r, ra, rb, and

rc. Let r, ra, rb, and rc be the first four integers

required; and let the three lengths of the sides of the triangle be the last three integers required.

Example 3

Pick the 5-12-13 right triangle. Then a = 5, b = 12, and c = 13. So

? 5 + 12-13 r

=-2

= 2,

r _ 5 - 12 + 13 a~

2

= 3,

? 12-5 + 13 rb

=-g

= 10, and

r _ 5 + 12 + 13 r?~ 2

= 15.

Thus, the seven integers are 2, 3,10,15, 5,12, and 13. Do these integers produce the correct

results? Of course:

2 + 3 + 10 = 15,

and

-a _ -q(q + ? + c) 6 + a

" (6 + c)(a+6+c)

_ -a2-ab-ac " (?+c)(a + ? + c)

_?2-c2-q(& + c) (6+c)(a + ?+c)

_ (?+c)(?-c)-(&+c)a (6+c)(q+6 + c)

_ (? + c)(-q + ?-c) (6+c)(a + ?+c)

_-a + ?-c ~ a + 6 + c

= equation (12)

Vol. 96, No. 5 ? May 2003 363



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22 + 32 + 102 + 152 = 52 + 122 + 132, 4 + 9 + 100 + 225 = 25 + 144 + 169,

338 = 338

Example 4

For the 7-24-25 right triangle, r = 3, ra = 4, rb = 21, and rc = 28. So the integers are 3,4, 21, 28, 7, 24, and 25.

Then

3 + 4 + 21 = 28, and

32 + 42 + 212 + 282 = 72 + 242 + 252, 9 + 16 + 441 + 784 = 49 + 576 + 625,

1250 = 1250.

The many beautiful results discovered here cer

tainly illustrate the pleasure to be derived from

studying mathematics. Only straightforward alge bra and some plane and analytic geometry are

needed to experience this pleasure.

REFERENCE Hansen, David W. "On The Radii of Inscribed and

Escribed Circles of Right Triangles." Mathematics Teacher 72 (September 1979): 462-64.

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