2009 USAMO Problems真题及答案

2009 USAMO Problems真题及答案

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Day 1

Problem 1

Given circles??and??intersecting at points??and?, let??be a line through the center of??intersecting??at points??and??and let??be a line through the center of??intersecting??at points??and?. Prove that if??and??lie on a circle then the center of this circle lies on line?.

Problem 2

Let??be a positive integer. Determine the size of the largest subset of??which does not contain three elements??(not necessarily distinct) satisfying?.

Problem 3

We define a?chessboard polygon?to be a polygon whose sides are situated along lines of the form??or?, where??and??are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping??rectangles. Finally, a?tasteful tiling?is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a??rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.

a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.

b) Prove that such a tasteful tiling is unique.

Day 2

Problem 4

For??let?,?, ...,??be positive real numbers such that

Prove that?.

Problem 5

Trapezoid?, with?, is inscribed in circle??and point??lies inside triangle?. Rays??and??meet??again at points?and?, respectively. Let the line through??parallel to??intersect??and??at points??and?, respectively. Prove that quadrilateral??is cyclic if and only if??bisects?.

Problem 6

Let??be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that??Suppose that??is also an infinite, nonconstant sequence of rational numbers with the property that??is an integer for all??and?. Prove that there exists a rational number??such that??and??are integers for all??and?.

[wechat keyword="usamo" key="usamohanlin"]

Problem 1

Solution

Let??be the?circumcircle?of?,??to be the radius of?, and??to be the center of the circle?, where?. Note that??and??are the?radical axises?of??,??and??,??respectively. Hence, by?power of a point(the power of??can be expressed using circle??and??and the power of??can be expressed using circle??and?),Subtracting these two equations yields that?, so??must lie on the?radical axis?of??,?.

~AopsUser101

Remarks

Conveniently, this analytic solution takes care of all configuration issues we may have encountered had we used a more traditional solution, such?as?angle-chasing(which would indeed work, just be considerably less elegant).

~AopsUser101

Problem 2

Solution 1

Let??be a subset of??of largest size satisfying??for all?. First, observe that?. Next note that?, by observing that the set of all the odd numbers in??works. To prove that?, it suffices to only consider even?, because the statement for??implies the statement for??as?well. So from here forth, assume??is even.

For any two sets??and?, denote by??the set?, and by??the set?. Also, let??denote??and??denote?. First, we present a lemma:

Lemma 1: Let??and??be two sets of integers. Then?.

Proof: Write??and??where??and?. Then??is a strictly increasing sequence of??integers in?.

Now, we consider two cases:

Case 1: One of??is not in?. Without loss of generality, suppose?. Let??(a set with??elements), so that??by our assumption. Now, the condition that??for all??implies that?. Since any element of??has absolute value at most?, we have?. It follows that?, so?. However, by Lemma 1, we also have?. Therefore, we must have?, or?, or?.

Case 2: Both??and??are in?. Then??and??are not in?, and at most one of each of the pairs??and their negatives are in?. This means??contains at most??elements.

Thus we have proved that??for even?, and we are done.

Solution 2

Let??be the set of subsets satisfying the??condition for?, and let??be the largest size of a set in?. Let??if??is even, and??if??is odd. We note that??due to the following constuction:or all of the odd numbers in the set. Then the sum of any three will be odd and thus nonzero.

Lemma 1:?. If?, then we note that?, so?.?

Lemma 2:?. Suppose, for sake of contradiction, that??and?. Remove??from?, and partition the rest of the elements into two sets?, where??and??contain all of the positive and negative elements of?, respectively. (obviously?, because?). WLOG, suppose?. Then?. We now show the following two sub-results:

Sub-lemma (A): if?,??[and similar for?]; andSub-lemma (B): we cannot have both??and??simultaneously hold.

This is sufficient, because the only two elements that may be in??that are not in??are??and?; for?, we must either have??and both??[but by pigeonhole?, see sub-lemma (A)], or?, and?, in which case by (A) we must have?, violating (B).

(A): Partition??into the??sets?. Because?, then if any of those sets are within?,?. But by Pigeonhole at most??elements may be in?, contradiction.

(B): We prove this statement with another induction. We see that the statement easily holds true for??or?, so suppose it is true for?, but [for sake of contradiction] false for?. Let?, and similarly for?. Again WLOG?. Then we have?.

If?, then by inductive hypothesis, we must have?. But (A) implies that we cannot add??or?. So to satisfy??we must have??added, but then??contradiction.

If?, then at least three of??added. But?, and by (A) we have that??cannot be added. If?, then another grouping similar to (A) shows that??canno be added, contradiction. So?,?, and adding the three remaining elements gives?contradiction.

If?, then all four of??must be added, and furthermore?. Then?, and by previous paragraph we cannot add?.?

So we have??and by induction, that?, which we showed is achievable above.

Problem 3

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Problem 4

Solution

Assume without loss of generality that?. Now we seek to prove that?.

By the?Cauchy-Schwarz Inequality,Since?, clearly?, dividing yields:

as?desired.

Problem 5

Solution 1

We will use directed angles in this solution. Extend??to??as?follows:

If:

Note thatThus,??is cyclic.

Also, note that??is cyclic becausedepending on the configuration.

Next, we have??are collinear since

Therefore,so??is cyclic.

Only If: These steps can be reversed.

Solution 2 (Projective)

Extend??to?, and let line??intersect??at??and another point?,?as?shown:

If:

Suppose that?, and?. Pascal's theorem on the tuple??implies that the points?,?, and??are collinear. However,??and??are symmetrical with respect to the axis of symmetry of trapezoid?, and??and??are also symmetrical with respect to the axis of symmetry of??(as??is the midpoint of?, and?). Since?,??and??are symmetric with respect to the axis of symmetry of trapezoid?. This implies that line??is equivalent to line?. Thus,??lies on line?. However,?, so this implies that?.

Now note that??is cyclic. Since?,?. However,?. Therefore,??is cyclic.

Only If:

Consider the same setup, except??is no longer the midpoint of?. Note that??must be parallel to??in order for??to be cyclic. We claim that??and hope to reach a contradiction. Pascal's theorem on the tuple??implies that?,?, and??are collinear. However, there exists a unique point??such that?,?, and??are concurrent. By?If,??must be the midpoint of??in order for the concurrency to occur; hence,?. Then?, since?. However, this is a contradiction, so therefore??cannot be parallel to??and??is not cyclic.

Solution by?TheBoomBox77

Problem 6

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