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USAMO 2007

Day 1Problem 1

Let be a positive integer. Define a sequence by setting and, for each, letting be the unique integer in the range for which

is divisible by . For instance, when the obtainedsequence is . Prove that for any the sequenceeventually becomes constant.

Solution Problem 2

A square grid on the Euclidean plane consists of all points , where andare integers. Is it possible to cover all grid points by an infinite family of discs withnon-overlapping interiors if each disc in the family has radius at least 5?

Solution

Problem 3

Let be a set containing elements, for some positive integer .Suppose that the -element subsets of are partitioned into two classes. Provethat there are at least pairwise disjoint sets in the same class.

Solution

Day 2Problem 4

An animal with cells is a connected figure consisting of equal-sized cells. Thefigure below shows an 8-cell animal.

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A dinosaur is an animal with at least 2007 cells. It is said to be primitive if its cellscannot be partitioned into two or more dinosaurs. Find with proof the maximumnumber of cells in a primitive dinosaur.Animals are also called polyominoes . They can be defined inductively. Two cells are adjacent if

they share a complete edge. A single cell is an animal, and given an animal with cells, one with

cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.Solution

Problem 5

Prove that for every nonnegative integer , the number is the product of atleast (not necessarily distinct) primes.

Solution

Problem 6

Let be an acute triangle with , , and being its incircle, circumcircle, andcircumradius, respectively. Circle is tangent internally to at and tangentexternally to . Circle is tangent internally to at and tangent internally to .Let and denote the centers of and , respectively. Define points ,

, , analogously. Prove that with equality if and only if triangle is equilateral.

Solution 1

By the above, we have that

, and by definition, . Thus, . Also, bothare integers, so . As the s form a non-increasing sequence of positiveintegers, they must eventually become constant.

Therefore, for some sufficiently large value of . Then, so eventually the sequence

becomes constant.

Solution 2Let . Since , we have that

.

Thus, .

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all sets in add that set to . The process terminates when every separablesubset intersects a set in . Let be the set of elements in which are not in anyset in . We claim that one class contains every -element subset of .

Suppose that are elements of . Denote by the set. Note that for each , is not separable, so that

and are in the same class. But then is in the same class for each— in particular, and are in the same class. But for

any two sets we may construct such a sequence with equal to one andequal to the other.

We are now ready to construct our disjoint sets. Suppose that . Then, so we may select disjoint -element

subsets of . Then for each of the sets in , we may select a subset which is inthe same class as all the subsets of , for a total of disjoint sets.

Solution 1Let a -dino denote an animal with or more cells.

We show by induction that an -dino with or more animal cells is notprimitive. (Note: if it had more, we could just take off enough until it had 4n-2,which would have a partition, and then add the cells back on.)

Base Case: If , we have two cells, which are clearly not primitive.Inductive Step: Assume any cell animal can be partitioned into two or more

-dinos.

For a given -dino, take off any four cells (call them ) to get ananimal with cells.

This can be partitioned into two or more -dinos, let's call them and . Thismeans that and are connected.

If both and are -dinos or if don't all attach to one of them,then we're done.

So assume has cells and thus has at least cells, and thatare added to . So has cells total.

Let denote the cell of attached to . There are cells on besides .Thus, of the three (or less) sides of not attached to , one of them must have

cells by the pigeonhole principle . It then follows that we can add , , andthe other two sides together to get an dino, and the side of that has

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cells is also an n-dino, so we can partition the animal with cells into two-dinos and we're done.

Thus, our answer is cells.

Example of a solution Attempting to partition solution intodinosaurs

Solution 2For simplicity, let and let be the number of squares . Let the centers of the squares be vertices , and connect any centers of adjacent squares withedges. Suppose we have some loops . Just remove an edge in the loop. We arestill connected since you can go around the other way in the loop. Now we haveno loops. Each vertex can have at most 4 edges coming out of it. For each point,assign it the quadruple : where , , , are the numbers of verticies oneach branch, WLOG . Note .

Claim: If , then we must be able to divide the animal into twodinosaurs. Chose a vertex, , for which is minimal (i.e. out of all maximal

elements in a quadruple, choose the one with the least maximal element). Wehave that , so . Hence we can just cut off thatbranch, that forms a dinosaur.

But suppose the remaining verticies do not make a dinosaur. Then we have. Now move to the

first point on the branch at . We have a new quadruple )where .

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Now consider the maximal element of that quadruple. We already have. WLOG , then

so , so isthe maximal element of that quadruple.

Also , so . But that is a contradiction to theminimality of . Therefore, we must have that , so we have apartition of two dinosaurs.

Maximum: . Consider a cross with each branch having verticies.Clearly if we take partition verticies, we remove the center, and we are notconnected.

So : .

Solution 3 (Generalization)

Turn the dinosaur into a graph (cells are vertices , adjacent cells connected by anedge) and prove this result about graphs. A connected graph with vertices,where each vertex has degree less than or equal to , can be partitioned into

connected components of sizes at least . So then in this special case, wehave , and so (a possible configuration of this size thatworks consists of a center and 4 lines of cells each of size 2006 connected to thecenter). We next throw out all the geometry of this situation, so that we have acompletely unconstrained graph. If we prove the above-mentioned result, we canput the geometry back in later by taking the connected components that our partition gives us, then filling back all edges that have to be there due to adjacentcells. This won't change any of the problem constraints, so we can legitimately dothis.

Going, now, to the case of arbitrary graphs, we WOP on the number of edges. If we can remove any edge and still have a connected graph, then we have found asmaller graph that does not obey our theorem, a contradiction due to theminimality imposed by WOP. Therefore, the only case we have to worry about iswhen the graph is a tree. If it's a tree, we can root the tree and consider the sizeof subtrees. Pick the root such that the size of the largest subtree is minimized.

This minimum must be at least , otherwise the sum of the size of thesubtrees is smaller than the size of the graph, which is a contradiction. Also, it

must be at most , or else pick the subtree of size greater than and you havedecreased the size of the largest subtree if you root from that vertex instead, so

you have some subtree with size between and . Cut the edge connectingthe root to that subtree, and use that as your partition.

It is easy to see that these partitions satisfy the contention of our theorem, so weare done.

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Solution 1

We proceed by induction.

Let be . The result holds for because is the product of primes.

Now we assume the result holds for . Note that satisfies the recursion

.

Since is an odd power of , is a perfect square. Thereforeis a difference of squares and thus composite , i.e. it

is divisible by primes. By assumption, is divisible by primes. Thus is divisible by primes as desired.

Solution 2

Notice that . Therefore it suffices

to show that is composite.

Let . The expression becomes

which is the shortened form of the geometric series . This can be factored as

.

Since is an odd power of , is a perfect square , and so we can factor this bydifference of squares. Therefore, it is composite.

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Lemma :

Proof :

Note and lie on since for a pair of tangent circles, the point of tangencyand the two centers are collinear .

Let touch , , and at , , and , respectively. Note. Consider an inversion , , centered at , passing through ,

. Since , is orthogonal to the inversion circle, so . Consider . Note that passes through and is tangent to , hence is a

line that is tangent to . Furthermore, because is symmetric about, so the inversion preserves that reflective symmetry. Since it is a line that is

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symmetric about , it must be perpendicular to . Likewise, isthe other line tangent to and perpendicular to .

Let and (second intersection).

Let and (second intersection).

Evidently, and . We want:

by inversion. Note that , and they are tangent to , so the distance between those lines is . Drop a perpendicular from to ,

touching at . Then . Then ,

= . So

Note that . Applying the double angle formulas and

, we get

End Lemma

The problem becomes:

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which is true because , equality is when the circumcenter and

incenter coincide. As before, , so, by symmetry,. Hence the inequality is true iff is equilateral.

Comment: It is much easier to determine by considering . We have, , , and . However, the inversion is

always nice to use. This also gives an easy construction for because thetangency point is collinear with the intersection of and .