Let $ ABCD$ be a convex quadrilateral, and let $ R_A, R_B, R_C, R_D$ denote the circumradii of the triangles $ DAB, ABC, BCD, CDA,$ respectively. Prove that $ R_A + R_C > R_B + R_D$ if and only if $ \angle A + \angle C > \angle B + \angle D.$
Problem
Source: IMO Shortlist 1996, G8
Tags: inequalities, trigonometry, geometry, circumcircle, IMO Shortlist
Bugi
09.08.2008 21:39
$ R_A = \frac {AD}{2sin\angle ABD}$ $ R_B = \frac {BC}{2sin\angle CAB}$ $ R_C = \frac {BC}{2sin\angle BDC}$ $ R_D = \frac {AD}{2sin\angle DCA}$ WLOG $ \angle DAB$ + $ \angle BCD$ $ > =$ $ \angle CDA$ + $ \angle ABC$ The sum of all angles is $ 360$ and that implies $ S =$ $ \angle DAB$ + $ \angle BCD$ $ > =$ $ 180$ 1)If $ S = 180$ then A,B,C and D lie on the same circle, so $ R_A + R_C = R_B + R_D$ 2)If $ S > 180$ then $ \angle DCA > \angle ABD$ and $ \angle CAB > \angle BDC$ which implies $ R_A > R_D$ and $ R_C > R_B$ Sum these two inequalities and you get the result
alkjash
16.09.2010 03:19
@Bugi: $\angle DCA > \angle ABD$ does not imply $\sin \angle DCA > \sin \angle ABD$ or the opposite inequality, since some of these angles may be obtuse.
By capital letters $A, B, C, D$, I mean those angles of the quadrilateral $ABCD$ with the specified point as vertex. Also, let $R(XYZ)$ denote the circumradius of triangle $XYZ$.
Clearly the if and only if statements are equivalent. Assume then that $A + C > B + D$ First we dispose of most quadrilaterals with one method, and then we look for those for which this method is invalid and prove the problem for them separately.
Assume WLOG $\sin C < \sin A$ so that $R_C > R_A$ by the law of sines. Extend $AC$ to meet the circumcircle of $ABD$ at point $C'$. We have $R_A + R_C > R_A + R(BC'D)$ since $R(BC'D) = R_A$, and $R_A + R(BC'D) = R(ABC') + R(ADC')$. Next, if $BC' > BC$ and $CD' > CD$, then we would have $R(ABC') = \frac{BC'}{\sin A} > \frac{BC}{\sin A} > R(ABC)$, and similarly $R(ADC') > R(ADC)$, so we would be done.
Hence it remains to consider $ABCD$ for which $BC' > BC$ and $CD' > CD$ are not both true. Clearly at least one of $\angle ACB$, $\angle ACD$ is acute, so at most one of these inequalities is false. WLOG, assume $BC' < BC$. Since the other inequality $CD' > CD$ holds we can show as before that $R_A > R_D$. We are left with proving $R_C > R_B$.
From $BC' < BC$ it follows that $\angle BCA$ is obtuse. Next, we show that $\angle BDC < \angle BAC$ and both are acute. The acuteness follows from the obtusenesses of $\angle BCA$ and $C$. The inequality is obvious from the fact that the ray $DC$ meets the circumcircle of $ABD$ at a point on minor arc $BC'$. Hence $\sin \angle BDC < \sin \angle BAC$ so $\frac{BC}{\sin \angle BDC} > \frac{BC}{\sin \angle BAC}$ and $R_C > R_B$.