THE ISOPERIMETRIC PROBLEM:

A PROPOSALBY: MIRIAM FELSENTHAL

DIDO’S PROBLEM• THERE IS A LEGEND IN THE AENEID OF QUEEN

DIDO.• SHE FLED TO NORTH AFRICA AND BARGAINED

WITH THE LOCAL RULER FOR A PLOT OF LAND THAT A BULL’S HIDE CAN COVER.• DIDO CUT THE HIDE INTO STRIPS AND

PROCEEDED TO LAY THEM OUT TO FORM A SEMI-CIRCLE, THE SHAPE THAT WOULD ENCOMPASS THE MOST TERRITORY, WHILE GAINING ACCESS TO THE SEA.• SHE APPARENTLY KNEW THE ANSWER TO THE

ISOPERIMETRIC PROBLEM.

DIDO’S PROBLEM

THE ISOPERIMETRIC PROBLEM• THE ISOPERIMETRIC PROBLEM IS THE CONCEPT

OF MAXIMIZING THE AREA WHILE MINIMIZING THE PERIMETER.• THROUGHOUT HISTORY, MANY

MATHEMATICIANS HAVE ENDEAVORED TO PROOF THAT IT IS THE CIRCLE OF ALL THE SHAPES OF EQUAL PERIMETER THAT HAS THE LARGEST AREA. • I PROPOSE TO EXPLORE THE GENERAL TOPIC

AND THEN TO USE MY RESEARCH TO IDENTIFY A NEW FACET TO THE ISOPERIMETRIC PROBLEM.

BASICS OF A CIRCLE

C = 2r

A =

ARCHIMEDES’ LOWER BOUND OF THE CIRCUMFERENCE

• PHEXAGON = 6 *

• PHEXAGON < PCIRCLE

• THUS, THE CIRCUMFERENCE OF A CIRCLE IS ALWAYS GREATER THAN THAT OF AN EQUILATERAL HEXAGON WHOSE SIDES EQUAL THE RADIUS OF THE CIRCLE.

r

• THE LINE THAT DICTATES THE SHAPE OF A SEMICIRCLE IS :• .

• TO EVALUATE THE AREA, IT IS NECESSARY TO TAKE THE INTEGRAL FROM ONE END OF THE SHAPE TO THE OTHER.

• THE SOLUTION TO THIS INTEGRAL IS:

• DOUBLING THIS FORMULA APPLIES IT TO A CIRCLE:

CALCULUS APPROACH

TWO FUNDAMENTALS OF THE ISOPERIMETRIC PROBLEM

• STATEMENT 1: AMONG ALL SHAPES OF THE SAME PERIMETER, THE CIRCLE HAS THE LARGEST AREA.• STATEMENT 2: AMONG ALL SHAPES OF THE

SAME AREA, THE CIRCLE HAS THE SMALLEST PERIMETER.

PROOF BY CONTRADICTION• ASSUME THAT STATEMENT 1 IS TRUE, BUT

STATEMENT 2 IS FALSE.

C F C’

• AC = AF • PC > PF

• PC’ = PF

• AC’ < AC → AC’ < AF

ZENODORUS

• GREEK MATHEMATICIAN FROM THE SECOND CENTURY BCE.• ON ISOPERIMETRIC FIGURES• THEON OF ALEXANDRIA AND PAPPUS (FOURTH

CENTURY CE)

• HE CLAIMED TO HAVE DEVELOPED A PROOF THAT THE SHAPE THAT IS EQUILATERAL AND EQUIANGULAR IS THE GREATEST OF ALL SHAPES THAT HAVE AN EQUAL NUMBER OF SIDES AND EQUAL PERIMETER.

FIRST LEMMA

• PSCALENE = PISOSCELES

• ACALENE < AISOSCELES

• IF AN ISOSCELES TRIANGLE AND A SCALENE TRIANGLE SHARE THE SAME BASE AND HAVE EQUAL PERIMETERS, THE AREA OF THE ISOSCELES TRIANGLE WILL BE LARGER.

SECOND LEMMA• WHEN THERE ARE TWO NON-SIMILAR ISOSCELES

TRIANGLES WITH A GIVEN SUMMED PERIMETER AND SUMMED AREA, IF ONE WERE TO CONSTRUCT SIMILAR ISOSCELES TRIANGLES ON THE RESPECTIVE BASES OF THE FIRST TWO SO THAT THE SUM OF THEIR PERIMETERS IS EQUAL TO THOSE OF THE ORIGINALS, THE SUM OF THE AREA OF THE SIMILAR TRIANGLES WILL BE GREATER THAN THAT OF THE NON-SIMILAR TRIANGLES.

C DA B

ZENODORUS’ PROOF: PART 1

• ACCORDING TO THE FIRST LEMMA, AAFC > AABC

C

D

A

B

E

F

• THUS, THE AREA OF THE PENTAGON WOULD BE LARGER IF ALL OF ITS SIDES WOULD BE EQUILATERAL.

• AF + FC = AB + BC

AND ABC IS A SCALENE ONE.

• AFC IS AN ISOSCELES TRIANGLE,

ZENODORUS’ PROOF: PART 2

• P ABC + CDE = P AFC + CGE.

C

D

AB

E

F

G

• ABC AND CDE ARE NON-SIMILAR ISOSCELES TRIANGLES.• DRAW AFC AND CGE SUCH THAT THEY ARE SIMILAR ISOSCELES TRIANGLES.

• ACCORDING TO THE SECOND LEMMA,

• A ABC + CDE < A AFC + CGE.• THUS, THE AREA OF THE PENTAGON WOULD BE LARGER IF ALL OF ITS SIDES WOULD BE EQUIANGULAR.

PAPPUS

• “BEES, THEN, KNOW JUST THIS FACT WHICH IS USEFUL TO THEM, THAT THE HEXAGON IS GREAT[EST]…AND WILL HOLD MORE HONEY FOR THE SAME EXPENDITURE OF MATERIAL IN CONSTRUCTING EACH. BUT WE, CLAIMING A GREATER SHARE IN WISDOM THAN THE BEES, WILL INVESTIGATE A SOMEWHAT WIDER PROBLEM, NAMELY THAT, OF ALL EQUILATERAL AND EQUIANGULAR PLANE FIGURES HAVING AN EQUAL PERIMETER, THAT WHICH HAS THE GREATER NUMBER OF ANGLES IS ALWAYS GREATER, AND THE GREATEST OF THEM ALL IS THE CIRCLE HAVING ITS PERIMETER EQUAL TO THEM.”

FURTHER EXPLORATION OF THE ISOPERIMETRIC PROBLEM

• PAPPUS PROPOSED THAT THE SEMI-CIRCLE WILL HAVE THE LARGEST AREA OF ALL CIRCULAR SEGMENTS THAT HAVE THE SAME CIRCUMFERENCE.• JAKOB STEINER, IN 1838, PRESENTED FIVE

PROOFS ON THE SUBJECT, YET THEY ALL ASSUME THE EXISTENCE OF A SOLUTION, WHICH RENDERS THEM UNSUITABLE AS RIGOROUS MATHEMATICAL PROOFS.• KARL WEIERSTRASS, IN 1879, FINALLY

PRESENTED A PROPER SOLUTION, THROUGH THE USE OF CALCULUS OF VARIATIONS.

ISOPERIMETRIC PROBLEM

• THE ISOPERIMETRIC PROBLEM DEALS WITH MAXIMIZING AREA AND MINIMIZING PERIMETER.• ISOPERIMETRIC INEQUALITY: • ISOPERIMETRIC QUOTIENT:• N-DIMENSIONS

MY PROPOSAL

• IN MY HONORS THESIS PAPER, I PLAN TO EXPLORE THESE TOPICS AND CULTIVATE A RICHER UNDERSTANDING AS TO THE MATHEMATICS BEHIND THIS HISTORICAL CHALLENGE.

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