Recent changes to Interactive Mathematics Miscellany and PuzzlesInteractive Mathematics Miscellany and Puzzles is a frequently updated site. Here are the most recent addtions. For other changes please visit the site.Sladjan Stankovik's Inequality With Constraint IIa+b+c+d=4. Prove that aaaa + bbbb + cccc + dddd + 4abcd ≥ 8
http://www.cut-the-knot.org/arithmetic/algebra/Stankovik2.shtml
An Inequality in Triangle and In Generalsum[(cot(A).cot^3(B))/(cot^2(B)+2cot^2(A))]+2sum[(cot^2(A)cot(B))/(cot(A)+2cot(B)) ≥ 1
http://www.cut-the-knot.org/m/Algebra/InequalityInTriangleAndInGeneral.shtml
Getting Ahead by Two PointsOne of the players wins if he gets ahead by two points. Playing 6 games, what is the expectation of the length of the competition
http://www.cut-the-knot.org/Outline/Probability/GettingAheadByTwoPoints.shtml
Radicals, Radicals, And More Radicals in an InequalityProve that, for 0 ≤ x≤ y ≤ z, sqrt(x)+sqrt(y)+sqrt(z) ≥ 2g.sqrt(3(x+y+z)), where g=(sqrt[4](xz))/(sqrt{x)+sqrt(z))
http://www.cut-the-knot.org/m/Algebra/LorianSaceanu.shtml
Find Minimum without CalculusFind Minimum without Calculus, f(x)=(5-4x+x^2)/(x-2), xless than 2
http://www.cut-the-knot.org/m/Calculus/MinimumWithoutCalculus.shtml
A Cyclic Inequality in Three Variables XIVsum (xy)/(xy+y^2+zx) ≤ 1
http://www.cut-the-knot.org/m/Algebra/CyclicInequalityInThreeVariables14.shtml
Isosceles Triangle with Two Points on BaseIn ann isosceles triangle BAC two points D,E on the bases are such that angle DAE is half angle BAC. Prove that BD, DE, EC may form a triangle and find the sum of a couple of angle
http://www.cut-the-knot.org/m/Geometry/IsoscelesTriangleWithTwoPointsOnBase.shtml
Angle Chasing in a 75-30-75 triangleIn a 75-30-75 triangle, we identify three equal equilateral triangles, four equal 45-90-45 triangles, and four equal 75-30-75 ones
http://www.cut-the-knot.org/m/Geometry/75-30-75.shtml
Nine Point Center in SquareOn a side of a square internally is built an equilateral triangle. The configuration contains seveal other triangles. Some four vertices form a rectangle
http://www.cut-the-knot.org/m/Geometry/NinePointCenterInSquare.shtml
An Inequality for the Cevians through Spieker Point via Brocard AngleAA', BB', CC' are the cevians via the Spieker point in triangle ABC. Then (a^2b^2+b^2c^2+c^2a^2) ≥ 2s(AC'\cdot BA'\cdot CB' + AB'\cdot BC'\cdot CA')
http://www.cut-the-knot.org/triangle/SpiekerViaBrocard.shtml