Solution　We have 4(10n+6) = 6.10^m + n, where n has m digits. So 13n + 8 = 2.10^m. Hence n = 2n‘ and 13n’ = 10^m - 4. Dividing, we quickly find that the smallest n‘, m satisfying this are: n’=7692, m=5. Hence the answer is 153846.

Solution　It is easy to show that the inequality implies |x-1| > Ö31/8, so x > 1 + Ö31/8, or x < 1 - Ö31/8. But the converse is not true.
Indeed, we easily see that x > 1 implies the lhs < 0. Also care is needed to ensure that the expressions under the root signs are not negative, which implies -1 <= x <= 3. Putting this together, suggests the solution is -1 <= x < 1 - Ö31/8, which we can easily check.

Solution　The rhombus WXCZ, where W is the midpoint of AB’, X is the midpoint of BC‘, and Z is the midpoint of BD’.

Solution　Put c = cos x, and use cos3x = 4c³ - 3c, cos 2x = 2 c² - 1. We find the equation given is equivalent to c = 0, c² = 1/2 or c² = 3/4. Hence x = pi/2, 3pi/2, pi/4, 3pi/4, pi/6, 5pi/6 or any multiple of pi plus one of these.

Solution　The key is to notice that if O is the center of the inscribed circle, then angle AOC = 270 - angle ABC (chase a few angles around and use the fact that opposite angles in a cyclic quadrilateral sum to 180). So O must be the intersection of the arc AC and the angle bisector of angle ABC. To prove the construction possible we use the fact that a quadrilateral ABCD has an inscribed circle iff AB + CD = BC + AD. For D near C on the circumcircle of ABC we have AB + CD < BC + AD, whilst for D near A we have AB + CD > BC + AD, so as D moves continuously along the circumcircle there must be a point with equality. [Proof that the condition is sufficient: it is clearly necessary (use fact that tangents from a point are of equal length). So take a circle touching AB, BC and AD and let the other tangent from C (not BC) meet AD in D‘. Then CD’ - CD = AD‘ - AD, hence D’=D.]

Solution Let the triangle be ABC with AB = AC, let the incenter be I and the circumcenter O. Let the distance IO be d, taking d positive if O is closer to A than I, negative if I is closer. Let the angle OAB be q.
Then r = (R + d) sin q, and r + d = R cos 2q. It helps to draw a figure to check that this remains true for the various possible configurations. Using cos 2q = 1 - 2 sin²q, we find that (d + R + r)(d² - R(R - 2r)) = 0. But OI < OA, so d is not - R - r. Hence result.

Solution First part is obvious. The wrong way to do the second part is to start looking for the locus of the center of a sphere which touches three edges. The key is to notice that the tangents to a sphere from a given point have the same length.
Let the tetrahedron be A₁A₂A₃A₄. Let S be the sphere inside the tetrahedron, S₁ the tetrahedron opposite A₁, and so on. Let the tangents to S from A_i have length a_i. Then the side A_iA_j has length a_i+a_j. Now consider the tangents to S₁ from A₁. Their lengths are a₁ + 2a₂, a₁ + 2a₃, and a₁ + 2a₄. Hence a₂ = a₃ = a₄. Similarly, considering S₂, we have that a₁ = a₃ = a₄.