 ## Page 2 of 5

32 At Right Angles | Vol. 3, No. 1, March 2014

Then he looked down at the mouth watering cake

that Mr. Jagirdar had ordered for him. Following

his gaze, Mr. Jagirdar, exclaimed: If my sons had

to divide the triangle so that each of them got

equal cake as well as equal lengths of that delicious

chocolate border, then I could have my cake and eat

it too!

Figure 2

But...he added gloomily, If I cut along the median

starting from the top to the midpoint of the longest

side and give the left portion to my older son, then

he is sure to protest that this is not a fair share since

the younger fellow has more of that chocolate line

matter that the areas are equal.

And if I measure out the chocolate line and cut in

such a way that they get equal amounts then they

will get unequal areas. I would hate to cause fights

simply because I did not make a fair division!

Mr. Z nodded and in true mathematical style, he

restated the problem succinctly: Given a triangle,

you want to know if there exists a line which bisects

the area as well as the perimeter of the triangle.

Mr. Jagirdar looked abashed at this – was the

problem which had troubled him for so long so

very easy to state? He had to agree that it was!

Well, said Mr. Z, if your plot was circular, square or

even rectangular, your solution would have been

easy. Yes, said Mr. Jagirdar, I remember enough

about high school geometry to see that for any of

these shapes, any line through the centre of the plot

would divide it along my specifications.

Mr. Z smiled and said, Yes; even if the land had

been in the shape of an equilateral or isosceles

triangle, your problem would be simpler. Your

sons would accept your decision as calmly as when

you divided a grilled sandwich for them by cutting

it along the median. Mr. Jagirdar frowned, the

conversation was getting too technical for him

triangle and an isosceles triangle and was briskly

explaining that the median to the base of these

would also be the perpendicular bisector and

voilà, the problem would be solved. Figure 3

shows Mr. Z’s sketch.

Figure 3

Suppose the given triangle ABC is isosceles and AB =

AC. Let D be the midpoint of BC. From fig. 3, we see

that AB + BD = AC + CD.

Also, area of triangle ABD = 1⁄2 BD x AD = 1⁄2 DC x

Mr. Jagirdar gloomily stabbed at the cake with

his knife. This scalene triangle makes life difficult!

Relax, said Mr. Z authoritatively. It is possible, and

I will tell you how to make the cut XY so that even

the most careful measurements will leave your sons

equally satisfied. And I will provide the justification

for the division if they want to prove that you have

been fair.

A

B

D

C ## Page 3 of 5

At Right Angles | Vol. 3, No. 1, March 2014 33

I must see this, said Mr. Jagirdar eagerly. And this is

the proof that Mr. Z provided him with:

Let BC = a, CA = b, AB = c and 2s = a + b + c. Let line

L cut the sides AB and AC at X and Y, respectively.

Suppose AX = x and AY = y. (See Fig. 4.)

Stage 1: Prove that Area of ∆AXY = 1⁄2 Area of

∆ABC if and only if 2xy = bc.

Area of ∆AXY = 1⁄2 xy sin A

Area of ∆ABC = 1⁄2 bc sin A

Substitution in the given expression gives

2xy = bc. The converse is proved similarly.

Stage 2: Show that XY bisects both the area and

the perimeter of the triangle if and only if the

simultaneous equations x + y = s, 2xy = bc have

real solutions.

If line XY bisects the area, then 2xy = bc; and

if XY bisects the perimeter of the triangle,

then perimeter of triangle AXY = perimeter of

quadrilateral XYCB. Substituting the given lengths

we get x + y = s. So if XY bisects the perimeter

and the area, then this set of equations has real

solutions. The converse is proved similarly.

Stage 3: Using these simultaneous equations

arrive at the condition that the roots are real if

s

2

− 2bc ≥ 0.

Arrive at the quadratic 2x2 − 2sx + bc = 0 and find

its discriminant.

Stage 4: Prove that s

2

− 2bc ≥ 0 is equivalent to

(s − b)2

+ (s − c)2

− (s − a)2 ≥ 0

Expand (s − b)2

+ (s − c)2

− (s − a)2

. This becomes

s

2 + b2 − 2bs + s

2 + c2 − 2cs − s

2 − a2 + 2as

= s

2 − 2s (b + c − a) + (b2 + c2 − a2

)

= s

2

− 2s (b + c − a) + ((b + c)

2 − a2 −2bc)

= s

2 − 2s(b + c − a) + (b + c + a) (b + c − a) −2bc

On substituting 2s = a + b + c we get s

2

− 2bc.

Stage 5: Prove that (s − b)2

+ (s − c)2

− (s − a)2 ≥ 0

in turn is equivalent to

(s − b)2

+ (a − c) b ≥ 0.

Expand (s − b)2

+ (s − c)2

− (s − a)2

using the

difference of squares formula for the last two

terms. The expression becomes

(s − b)2 + (s − c − s + a)(s − c + s − a). Now

substitute 2s = a + b + c. The expression now

Figure 4 ## Page 4 of 5

34 At Right Angles | Vol. 3, No. 1, March 2014

reduces to (s − b)2

+ (a − c) b which is positive

since BC is the longest side

Stage 6: Explain why this expression ensures that

The discriminant being non-negative ensures that

Stage 7: Prove that the farmer can be successful

in his endeavor if he takes

x = s ∓ (s 2 − 2bc)

2

y= s ∓ (s 2 − 2bc)

2

and

x = s ∓ (s 2 − 2bc)

2

y= s ∓ (s 2 − 2bc)

2

Find the roots of the quadratic equation using the

formula.

Awesome said Mr. Jagirdar. I remember

memorizing the conditions for the nature of the

roots of a quadratic! I never thought that I would

rely them on them to bring peace among my sons!

That’s the problem, said Mr. Z cryptically.

Teacher Note:

Introduction: This problem may seem intimidating at first sight. But it is based on simple principles

learnt at school and can be solved by students if they are given a little scaffolding (see blue text for

suggested scaffolding and green text for answers which students may be able to provide).

Prior Knowledge:

1. Basic trigonometry

2. Formula for area of a triangle in terms of 2 sides and the included angle.

3. Concept of median of a triangle.

4. The discriminant and the nature of the roots of a quadratic.

Suggestion: Discuss with the students, the seeming contradiction between the diagram which seems

to suggest that x ≠ y and the result itself which seems to indicate the opposite.

Explain that the positive square root for x coincides with the negative square root for y and vice-versa.

PRITHWIJIT DE is a member of the Mathematical Olympiad Cell at Homi Bhabha

Centre for Science Education (HBCSE), TIFR. He loves to read and write popular

articles in mathematics as much as he enjoys mathematical problem solving.

His other interests include puzzles, cricket, reading and music. He may be con- tacted at de.prithwijit@gmail.com.

SNEHA TITUS is Associate Editor of At Right Angles and Assistant Professor and

Mathematics resource person at the University Resource Centre.

SWATI SIRCAR is Senior Lecturer and Resource Person at the University

Resource Centre of Azim Premji University. Math is the 2nd love of her life (the

first being drawing). She is a B.Stat-M.Stat from Indian Statistical Institute and

an MS in math from University of Washington, Seattle. She has been doing

mathematics with children and teachers for 5 years and is deeply interested in

anything hands on - origami in particular. 