Bretschneider's Theorem and Cyclic Quadrilaterals
Date: 11/30/2000 at 12:44:32 From: Rachel Blumberg Subject: I need to prove Bretschneider's theorem My Calculus III professor brought up this problem in class: Prove that maximizing any quadrilateral with sides ABCD means inscribing it into a circle. I used Bretschneider's theorem, which states that if s = (a+b+c+d)/2 and A and C are opposing angles in the quadrilateral, then the area of the quadrilateral is sqrt((s-a)(s-b)(s-c)(s-d)-((cos((A+C)/2)^2)). When this quadrilateral was inscribed, I saw that the opposing angles would be supplementary, and therefore the area would be maximized because the cosine term would be 0. My professor patted me on the back and said, "Now prove the theorem. And while you're at it, explain why defining a quadrilateral with supplementary opposing angles means that it is inscribable into a circle." Any insight on proving this theorem would be helpful. I do not know where to start. Also I know that a quadrilateral inscribed in a circle has opposing angles that add to 180 degrees, but I don't know how to show the converse, that if a quadrilateral has supplementary opposing angles, then it can be placed inside a circle.
Date: 11/30/2000 at 16:58:43 From: Doctor Rob Subject: Re: I need to prove Bretschneider's theorem Thanks for writing to Ask Dr. Math, Rachel. I once proved the cyclicity of the maximal-area quadrilateral with given side lengths by setting up the area as the sum of the areas of the two triangles formed by a diagonal, and then taking the derivative with respect to the length of that diagonal. When I set the derivative equal to zero and solved, I found that the length of the diagonal was such that the circumcircles of the two triangles coincided, so the quadrilateral was cyclic. I do not recommend doing this, however, since the computations were extraordinarily messy. Using Bretschneider's Formula was a good idea. Proving that may turn out to be messy. I don't know what Bretschneider's original proof was. To show that when opposite angles are supplementary the quadrilateral is cyclic is not too hard. Pick one of the two angles, say <ABC, and draw the diagonal AC. This forms a triangle ABC. Draw the circumcircle of this triangle. Then the arc AC has measure double that of of <ABC. Now consider line CD. This intersects the circle at point E. Draw EA. Then <CEA has measure half that of arc CA. Since arcs CA and AC have total measure 2 pi rad, or 360 degrees, <CEA and <ABC have total measure pi rad, or 180 degrees. Now since <ABC and <CDA are supplementary, and <ABC and <CEA are also supplementary, then <CEA = <CDA. But since the angles of triangle CDE add up to pi rad or 180 degrees, this implies that points E and D coincide, and so D lies on the circle. (This uses the fact that any exterior angle of a triangle is the sum of the two other interior angles.) Now for Bretschneider's Formula. Let the side lengths be a, b, c, and d. Let alpha be the measure of the angle between sides a and b, and beta be the measure of the angle between sides c and d. Now the areas of the two triangles formed by sides a, b, and one diagonal, and sides c, d, and the same diagonal, are K1 = (1/2)*a*b*sin(alpha) K2 = (1/2)*c*d*sin(beta) Then the area of the quadrilateral is K = K1 + K2 As you stated, we want s = (a+b+c+d)/2 and then we want to show that K = sqrt[(s-a)*(s-b)*(s-c)*(s-d)] - a*b*c*d*cos(alpha+beta) To eliminate K1 and K2 isn't bad, just use substitution. K = a*b*sin[alpha]/2 + c*d*sin[beta]/2 I'm not quite sure how to proceed from here. Good luck. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Date: 12/01/2000 at 13:07:06 From: Doctor Rob Subject: Re: I need to prove Bretschneider's theorem Hi, Rachel. I believe I have the proof of Bretschneider's Theorem now. Cut the quadrilateral into two triangles using one of the diagonals. Let the two angles whose vertices are not on that diagonal have measures alpha and beta, let the lengths of the sides adjacent to angle alpha be a and b, let the lengths of the sides adjacent to angle beta be c and d, and let the length of the diagonal be p. Then the area of the quadrilateral is the sum of the areas of the triangles. One triangle has base a and height b*sin(alpha), and the other has base c and height d*sin(beta). Thus the area of the quadrilateral is given by K = (1/2)*a*b*sin(alpha) + (1/2)*c*d*sin(beta) Square this expression: K^2 = (1/4)*(a^2*b^2*sin^2[alpha]+ 2*a*b*c*d*sin[alpha]*sin[beta]+ c^2*d^2*sin^2[beta]) Now use the Law of Cosines on each triangle: p^2 = a^2 + b^2 - 2*a*b*cos(alpha) p^2 = c^2 + d^2 - 2*c*d*cos(beta) Set these equal to each other to eliminate the variable p, and isolate the cosines on one side of the equation: 2*a*b*cos(alpha) - 2*c*d*cos(beta) = a^2 + b^2 - c^2 - d^2 Now square both sides of this equation, divide by 16, and bring all terms to the left-hand side: (1/4)*(a^2*b^2*cos^2[alpha]- 2*a*b*c*d*cos[alpha]*cos[beta]+ c^2*d^2*cos^2[beta]) - (a^2+b^2-c^2-d^2)^2/16 = 0 Add this to the equation for K^2 above, and use the fact that sin^2(x) + cos^2(x) = 1, once with x = alpha and once with x = beta: K^2 = (1/4)*(a^2*b^2+c^2*d^2+ 2*a*b*c*d*[sin(alpha)*sin(beta)-cos(alpha)*cos(beta)]) - (a^2+b^2-c^2-d^2)^2/16 Now it is a matter of simplifying: K^2 = (1/4)*(a^2*b^2+c^2*d^2-2*a*b*c*d*cos[alpha+beta]) - (a^2+b^2-c^2-d^2)^2/16 = (4*a^2*b^2+4*c^2+d^2-[a^2+b^2-c^2-d^2]^2)/16 - a*b*c*d*cos(alpha+beta)/2 = 2*a^2*b^2+2*a^2*c^2+2*a^2*d^2+2*b^2*c^2+2*b^2*d^2+2*c^2*d^2 - a^4-b^4-c^4-d^4)/16 - a*b*c*d*cos(alpha+beta)/2 Now the numerator of the first fraction almost factors, and it will if we add and subtract 8*a*b*c*d: K^2 = (-a+b+c+d)*(a-b+c+d)*(a+b-c+d)*(a+b+c-d)/16 - a*b*c*d/2 - a*b*c*d*cos(alpha+beta)/2 Here we introduce s = (a+b+c+d)/2. K^2 = (s-a)*(s-b)*(s-c)*(s-d) - a*b*c*d/2 - a*b*c*d*cos(alpha+beta)/2 = (s-a)*(s-b)*(s-c)*(s-d) - a*b*c*d*(1+cos[alpha+beta])/2 = (s-a)*(s-b)*(s-c)*(s-d) - a*b*c*d*cos^2([alpha+beta]/2) by using the cosine half-angle trigonometric identity. Finally, K = sqrt[(s-a)*(s-b)*(s-c)*(s-d) - a*b*c*d*cos^2([alpha+beta]/2)] This last is Bretschneider's Formula, so we have proved what we set out to. It's a bit complicated, but nothing really hard is involved (with the possible exception of noticing that a certain polynomial would factor if we added and subtracted the right thing). - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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