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1. A complete chess board contains an equal number of black and white squares. The two opposite corner squares have the same color, so if they're removed, the new shape contains different numbers of black and white squares. But every domino contains exactly one black and one white square, so after cutting out 30 dominos from the new shape, you are left with two squares of the same color that are therefore not adjacent.
2. Color the bowls alternately red and blue so that the bowl initially containing two bowls is red. Let R denote the number of beans in the red bowls, and let B denote the number of beans in the blue bowls. Then every move increases both R and B by one, so R-B always remains constant, no matter what you do. But in the starting configuration, R-B is 2, and in an end configuration, R-B is 0. So you can't reach an end configuration.
3.
a) Let s_a, s_b, and s_c denote the three angle bisectors of the triangle. Then s_a is the location of all the points inside the triangle with equal distance to b and c, and so on.
Let d(P,g) denote the distance between a point P and a line g.
Now let I denote the intersection of s_b and s_c. I lies on s_b, so d(I,c) = d(I,a). But I also lies on s_c, so d(I,c) = d(I,a) = d(I_b). Therefore I also lies on s_a and is therefore the common point of s_a, s_b, and s_c.
b) Let D, E and F be the midpoints of BC, CA and AB. Then AF = FB, BD = DC, CE = EA. Then AF/FB * BD/DC * CE/EA = 1*1*1 = 1, and
Ceva's theorem says that in this case AD, BE and CF intersect in one point.
c) Let AD, BE and CF denote the three altitudes of the triangle.
Thales' theorem says that ABDE, BCEF and CAFD are cyclic quadrilaterals. The
inscribed angle theorem now says that <EDA = <EBA = <EBF = <ECF = <ACF = <ADF, therefore DA bisects <EDF. Similarly, EB bisects <FED and FC bisects <DFE. So AD, BE and CF are the angle bisectors of DEF. They intersect, see part 3a).
4. Let p be the uniform scaling mapping k' onto k, so that p(C) = C. Now p(AB) is parallel to AB, and also a tangent to k in p(D). Therefore |p(D)A| = |p(D)b|. Using the inscribed angle theorem, it follows that <DCB = <p(D)CB = <ACp(D) = <ACD, therefore CD bisects ACB.
5. Let M be the set of configurations C of 1000 segments that have a red and a blue end each so that every point is an end of a segment in C, let D be the one with the shortest total segment length. Let us assume that there exist two segments PR and QS of D that intersect in T, where P and Q are blue and R and S are red. Now let D' be the configuration that is identical to D, except that D' contains PS and QR instead of PR and QS. Now D' is also in M, and the total segment length of D' is smaller than that of D, which is a contradiction to the assumption that D is the "shortest" configuration in M.
Therefore D can have no intersections.
6. Let us assume that you bought some gumballs but don't have three of the same color. Then you must have at most two of every color, therefore you cannot have bought more than 10 gumballs.
Therefore it follows that if you buy 11 gumballs, you must have three of the same color.
7. Let us define 100 groups labeled 1, 3, 5, 7, ..., 199. Let us divide the set of integers from 1 to 200 into those groups according to their largest odd factor. So if a number n is in a group labeled g, then n is equal to g times a power of 2. So if you pick two numbers out of the same group, then their quotient is a power of 2, therefore one is a multiple of the other.
Now if you pick 101 distinct integers between 1 and 200, you are bound to pick two who are in the same group. Of those two, one is a multiple of the other.
Teaser question: Come on, discuss it already.
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