aa r X i v : . [ m a t h . G M ] J a n AN EXTENSION OF PYTHAGORAS THEOREM
Mircea Gotea ∗ Abstract.
This article proves a Pythagoras-type formula for the sides and diagonals of apolygon inscribed in a semicircle having one of the sides of the polygon as diameter.
Keywords and phrases: cyclic polygon, Pythagoras theoremMSC 2010: 51N20 T his paper presents a Pythagoras type theorem regarding a cyclic polygon having one of thesides as a diameter of the circumscribed circle.Let us first observe that Pythagoras theorem [1–5] and its reciprocal can be stated differently. Theorem 1.
If a triangle is inscribed in a semicircle with radius R , such that its longestside is the diameter of the semicircle, then the following relation exists between its sides: a = b + c , (1) which can also be expressed as R = b + c . (2) Its reciprocal can be states as follows: if the relation a = b + c exists between the sides of atriangle, then that triangle can be inscribed in a semicircle such that the longest side is alsothe diameter of the semicircle. We can state and prove a similar theorem for quadrilaterals. Let
ABCD be a quadrilateralwith vertices B and C located on the semicircle that has the longest side AD as diameter. ∗ Retired teacher of Mathematics, Gurghiu, Mureş County, Romania.E-mail: [email protected] n Extension of Pythagoras Theorem da b cA B C D Theorem 2.
If a quadrilateral
ABCD with its longest side AD = d can be inscribed in asemicircle with the diameter AD , then its sides satisfy the following relation: d = a + b + c + 2 abcd . (3) Proof.
We can successively deduce the following: AD = AB + BD = AB + BC + CD − BC · CD · cos( C )= AB + BC + CD + 2 BC · CD · cos( A )= AB + BC + CD + 2 BC · CD · ABAD . (cid:3)
Particular case. If B coincides with A , then a = 0 , the quadrilateral becomes triangle ACD , and relation (3) becomes d = b + c , which represents Pythagoras theorem for thisright triangle.The reciprocal of Theorem 2 does not hold true: if the sides of a quadrilateral ABCD satisfy relation (3), the quadrilateral does not have to be inscribed in a semicircle havingthe longest side as diameter. For example, a quadrilateral with sides AD = 4 √ , AB = √ , BC = 3 + √ , CD = 3 − √ and m ( ∢ A ) = 90 ◦ does satisfy relation (3), has AD as itslongest side, but cannot be inscribed in a semicircle with diameter AD .However, the following statement is true: Lemma.
Let four segments have the lengths a , b , c , and d . If they satisfy relation (3), thenthere is at least one quadrilateral with these sides that can be inscribed in a semicircle havingthe longest side as diameter. ircea Gotea Proof.
On a circle with diameter AD = d = 2 R , we set a point B such that AB = a , and apoint C such that BC = b . We can prove that CD = c when the sides satisfy relation (3).We have successively: CD = AD cos( D )= − AD cos( B )= − AD · AB + BC − AC · AB · BC = − AD · AB + BC − ( AD − CD )2 · AB · BC = − d · a + b + c − d ab = c. We thus found a quadrilateral with sides of lengths a, b, c, d .Let us also observe that if a, b , and c are distinct, there are three incongruent quadrilateralsthat satisfy this requirement, depending on which of the three sides, a, b , or c , is selected asthe opposite side to d . (cid:3) Theorem 3.
If a pentagon
ABCDE with sides AB = a , BC = b , CD = c , DE = d , AE = 2 R is inscribed in a circle with radius R , then its sides satisfy the following relation: R = a + b + c + aby + xcdR , (4) where x = AC and y = CE . Ra b c dyxA B C D E n Extension of Pythagoras Theorem Proof.
In the quadrilateral
ABCE we can apply relation (3): R = a + b + y + abyR . (5)Using the Law of Cosines we have y = c + d − cd cos( D )= c + d + 2 cd cos ( ∢ CAE )= c + d + 2 cd x R .
Replacing y from this relation in (5), we obtain relation (4). (cid:3) For a hexagon, we can similarly prove the following:
Theorem 4.
If we inscribe a hexagon in a semicircle such that the longest side is also thediameter, then the following relation is true: R = a + b + c + d + e + abz + ycx + udeR . (6) Ra b c d exuy zA B C D E F
We can restate the theorem in the case of a polygon with n sides. Theorem 5.
If a polygon A A A . . . A n − A n can be inscribed in a semicircle with diameter A A n , then the following relation is true: ( A A n ) = n − X k =1 ( A k A k +1 ) + 2 n − X k =1 ( A A k +1 ) ( A k +1 A k +2 ) ( A k +2 A n ) A A n . (7) Observation.
The points A , A k +1 , A k +2 , and A n , for ≤ k ≤ n − , involved in the sumfrom the right side of formula (7), are the vertices of a cyclic quadrilateral inscribed in asemicircle. ircea Gotea Proof.
By using the mathematical induction technique, we have already demonstrated theparticular cases for n = 3 and n = 4 .Let us assume that the property is true for any polygon with n sides and let us considerthe case of a polygon A A A . . . A n A n +1 , for n ≥ , inscribed in a circle with diameter A A n +1 . Following the induction hypothesis, in an n -sided polygon A A A . . . A n − A n +1 the following relation is true: ( A A n +1 ) = n − X k =1 ( A k A k +1 ) + ( A n − A n +1 ) (8) +2 n − X k =1 ( A A k +1 ) ( A k +1 A k +2 ) ( A k +2 A n +1 ) A A n +1 . Applying the Law of Cosines in triangle A n − A n A n +1 , and using the facts that: m ( ∢ A n − A n A n +1 ) + m ( ∢ A n − A A n +1 ) = 180 ◦ and m ( ∢ A A n − A n +1 ) = 90 ◦ , we obtain: ( A n − A n +1 ) = ( A n − A n ) + ( A n A n +1 ) − A n − A n ) ( A n A n +1 ) cos ( A n )= ( A n − A n ) + ( A n A n +1 ) + 2 ( A n − A n ) ( A n A n +1 ) cos ( ∢ A n − A A n +1 )= ( A n − A n ) + ( A n A n +1 ) + 2 ( A n − A n ) ( A n A n +1 ) · A A n − A A n +1 . Substituting ( A n − A n +1 ) into relation (8), we obtain: ( A A n +1 ) = n − X k =1 ( A k A k +1 ) + ( A n − A n ) + ( A n A n +1 ) + 2 ( A n − A n ) ( A n A n +1 ) · A A n − A A n +1 +2 n − X k =1 ( A A k +1 ) ( A k +1 A k +2 ) ( A k +2 A n +1 ) A A n +1 = n X k =1 ( A k A k +1 ) + 2 n − X k =1 ( A A k +1 ) ( A k +1 A k +2 ) ( A k +2 A n +1 ) A A n +1 . The proof is now complete. (cid:3) n Extension of Pythagoras Theorem We can also note the validity of a reciprocal of this theorem: Let A A , A A , . . . , A n − A n , A n A be n segments with lengths A A = a , A A = a , . . . , A n − A n = a n − , A n A = a n .If these segments satisfy relation (7), then there is at least one polygon with these segmentsas sides that can be inscribed in the semicircle with diameter A A n . The proof can bedeveloped in the same manner as for the case of n = 4 . Acknowledgments:
I would like to thank Professor Aurel I. Stan, Department of Math-ematics, The Ohio State University, U.S.A., for his kind advice and encouragements.
Note:
This article was originally published in Romanian here [6].
References [1] E. Rusu.
De la Tales la Einstein . Editura Albatros, 1971.[2] D.B. Wagner.
A Proof of the Pythagorean Theorem by Liu Hui (Third Century A.D.).
In:
Historia Mathematica
12 (1985), pp. 71–73.[3] J. Hadamard.
Lessons in geometry. I. Plane geometry . Trans. by M. Saul. AmericanMathematical Society, 2008.[4] O. Bottema.
Topics in elementary geometry . Springer-Verlag, New York, 2008.[5] M.P. Saikia.
The Pythagoras Theorem.
In:
Asia Pacific Mathematics Newsletter
O extindere a teoremei lui Pitagora.
In: