MathDB

3

Part of 1993 IMO Shortlist

Problems(3)

American Shortlist Inequality

Source: IMO Shortlist 1993, USA 3

10/24/2005
Prove that ab+2c+3d+bc+2d+3a+cd+2a+3b+da+2b+3c≥23 \frac{a}{b+2c+3d} +\frac{b}{c+2d+3a} +\frac{c}{d+2a+3b}+ \frac{d}{a+2b+3c} \geq \frac{2}{3} for all positive real numbers a,b,c,da,b,c,d.
inequalitiesfunctionfour variables4-variable inequalityalgebraIMO Shortlist
Orthic triangle inequality

Source: IMO Shortlist 1993, Canada 2

3/15/2006
Let triangle ABCABC be such that its circumradius is R=1.R = 1. Let rr be the inradius of ABCABC and let pp be the inradius of the orthic triangle A′B′C′A'B'C' of triangle ABC.ABC. Prove that p≤1−13⋅(1+r)2. p \leq 1 - \frac{1}{3 \cdot (1+r)^2}. [hide="Similar Problem posted by Pascual2005"]
Let ABCABC be a triangle with circumradius RR and inradius rr. If pp is the inradius of the orthic triangle of triangle ABCABC, show that pR≤1−(1+rR)23\frac{p}{R} \leq 1 - \frac{\left(1+\frac{r}{R}\right)^2}{3}.
Note. The orthic triangle of triangle ABCABC is defined as the triangle whose vertices are the feet of the altitudes of triangle ABCABC.
SOLUTION 1 by mecrazywong:
p=2Rcos⁡Acos⁡Bcos⁡C,1+rR=1+4sin⁡A/2sin⁡B/2sin⁡C/2=cos⁡A+cos⁡B+cos⁡Cp=2R\cos A\cos B\cos C,1+\frac{r}{R}=1+4\sin A/2\sin B/2\sin C/2=\cos A+\cos B+\cos C. Thus, the ineqaulity is equivalent to 6cos⁡Acos⁡Bcos⁡C+(cos⁡A+cos⁡B+cos⁡C)2≤36\cos A\cos B\cos C+(\cos A+\cos B+\cos C)^2\le3. But this is easy since cos⁡A+cos⁡B+cos⁡C≤3/2,cos⁡Acos⁡Bcos⁡C≤1/8\cos A+\cos B+\cos C\le3/2,\cos A\cos B\cos C\le1/8.
SOLUTION 2 by Virgil Nicula:
I note the inradius r′r' of a orthic triangle.
Must prove the inequality r′R≤1−13(1+rR)2.\frac{r'}{R}\le 1-\frac 13\left( 1+\frac rR\right)^2.
From the wellknown relations r′=2Rcos⁡Acos⁡Bcos⁡Cr'=2R\cos A\cos B\cos C
and cos⁡Acos⁡Bcos⁡C≤18\cos A\cos B\cos C\le \frac 18 results r′R≤14.\frac{r'}{R}\le \frac 14.
But 14≤1−13(1+rR)2⟺13(1+rR)2≤34⟺\frac 14\le 1-\frac 13\left( 1+\frac rR\right)^2\Longleftrightarrow \frac 13\left( 1+\frac rR\right)^2\le \frac 34\Longleftrightarrow
(1+rR)2≤(32)2⟺1+rR≤32⟺rR≤12⟺2r≤R\left(1+\frac rR\right)^2\le \left(\frac 32\right)^2\Longleftrightarrow 1+\frac rR\le \frac 32\Longleftrightarrow \frac rR\le \frac 12\Longleftrightarrow 2r\le R (true).
Therefore, r′R≤14≤1−13(1+rR)2⟹r′R≤1−13(1+rR)2.\frac{r'}{R}\le \frac 14\le 1-\frac 13\left( 1+\frac rR\right)^2\Longrightarrow \frac{r'}{R}\le 1-\frac 13\left( 1+\frac rR\right)^2.
SOLUTION 3 by darij grinberg:
I know this is not quite an ML reference, but the problem was discussed in Hyacinthos messages #6951, #6978, #6981, #6982, #6985, #6986 (particularly the last message).
inequalitiesgeometrycircumcircleinradiustrigonometryIMO Shortlistgeometric inequality
representation of the number a in the base b

Source: IMO Shortlist 1993, Romania 2, created by Radu Todor

3/24/2006
Let a,b,na,b,n be positive integers, b>1b > 1 and bn−1∣a.b^n-1\mid a. Show that the representation of the number aa in the base bb contains at least nn digits different from zero.
number theoryrepresentationDivisibilityIMO Shortlist