A convex ABCDE is inscribed in a unit circle, AE being its diameter. If AB=a, BC=b, CD=c, DE=d and ab=cd=14, compute AC+CE in terms of a,b,c,d.
Problem
Source: Turkey TST 1997 Problem 4
Tags: trigonometry, geometry proposed, geometry
02.12.2011 20:18
AC + CE = -1/4 + 1/2 sqrt[(257/16) - 4(a^2 + b^2)] + 1/2 sqrt[(257/16) - 4(c^2 + d^2)].
15.07.2012 05:00
Vo Duc Dien wrote: AC + CE = -1/4 + 1/2 sqrt[(257/16) - 4(a^2 + b^2)] + 1/2 sqrt[(257/16) - 4(c^2 + d^2)]. No, it is really not. AC+CE=ad(√64a2−1+√64d2−1)+d(√4−a2+√4−d2)8ad which is not so nice. I don't know the answer. But I observed some good points about the construction. The distance of B to AC is 18. The distance of D to CE is 18. To construct such a pentagon, AC/2<1−1/8=7/8 and CE/2<1−1/8=7/8. AC<7/4 and CE<7/4. 28∘<cos−1(7/8)<∠CAE<sin−1(7/8)<62∘. If a is given, it is possible to find b,c,d in terms of a. 7+√154<AC+CE<2√2 and according to Geometer's Sketchpad AC+CE≈a+b+c+d−110.
04.05.2013 09:33
Let AC=x and CE=y. We have BE2=4−a2, AD2=4−d2, and x2+y2=4. Ptolemy in ABCE: (ay+2b)2=x2(4−a2) a2y2+4b2+4aby=4x2−a2x2 Ptolemy in ACDE: (2c+dx)2=y2(4−d2) d2x2+4c2+4cdx=4y2−d2y2 Adding these two side by side 4b2+4c2+a2y2+a2x2+d2x2+d2y2+4aby+4cdx=4x2+4y2 substitute x2+y2=4, ab=cd=1/4 4a2+4b2+4c2+4d2+x+y=16 x+y=16−4a2−4b2−4c2−4d2=4(4−a2−b2−c2−d2).◼
01.03.2021 00:45
From the law of Cos and ab=cd=14, |AC|2=a2+b2−12cosB, |CE|2=c2+d2−12cosD, Then, 4=a2+b2+c2+d2−12(cosB+cosD) From the triangle ACE, −cosB=|CE|2 and −cosD=|AC|2,so we get: 4=a2+b2+c2+d2+14(|AC|+|CE|), |AC|+|CE|=16−4(a2+b2+c2+d2)=4[4−(a2+b2+c2+d2)]