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[Solutions] United States of America TST Selection Test 2014

  1. Let $\leftarrow$ denote the left arrow key on a standard keyboard. If one opens a text editor and types the keys "$ab\leftarrow cd \leftarrow \leftarrow e \leftarrow \leftarrow f$", the result is "$faecdb$". We say that a string $B$ is reachable from a string $A$ if it is possible to insert some amount of $\leftarrow$'s in $A$, such that typing the resulting characters produces $B$. So, our example shows that "faecdb" is reachable from "abcdef". Prove that for any two strings $A$ and $B$, $A$ is reachable from $B$ if and only if $B$ is reachable from $A$.
  2. Consider a convex pentagon circumscribed about a circle. We name the lines that connect vertices of the pentagon with the opposite points of tangency with the circle gergonnians.
    a) Prove that if four gergonnians are conncurrent, the all five of them are concurrent.
    b) Prove that if there is a triple of gergonnians that are concurrent, then there is another triple of gergonnians that are concurrent.
  3. Find all polynomials $P(x)$ with real coefficients that satisfy \[P(x\sqrt{2})=P(x+\sqrt{1-x^2})\]for all real $x$ with $|x|\le 1$.
  4. Let $P(x)$ and $Q(x)$ be arbitrary polynomials with real coefficients, and let $d$ be the degree of $P(x)$. Assume that $P(x)$ is not the zero polynomial. Prove that there exist polynomials $A(x)$ and $B(x)$ such that
    • both $A$ and $B$ have degree at most $d/2$,
    • at most one of $A$ and $B$ is the zero polynomial,
    • $\dfrac{A(x)+Q(x)B(x)}{P(x)}$ is a polynomial with real coefficients. That is, there is some polynomial $C(x)$ with real coefficients such that $$A(x)+Q(x)B(x)=P(x)C(x).$$
  5. Find the maximum number $E$ such that the following holds: there is an edge-colored graph with 60 vertices and $E$ edges, with each edge colored either red or blue, such that in that coloring, there is no monochromatic cycles of length 3 and no monochromatic cycles of length 5.
  6. Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*} ca &- db \\ ca^2 &- db^2 \\ ca^3 &- db^3 \\ ca^4 &- db^4 \\ &\vdots \end{align*} and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$.

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Ả-rập Xê-út,1,Abel,5,Albania,2,AMM,2,Amsterdam,5,Ấn Độ,1,An Giang,21,Andrew Wiles,1,Anh,2,Áo,1,APMO,19,Ba Đình,2,Ba Lan,1,Bà Rịa Vũng Tàu,51,Bắc Giang,49,Bắc Kạn,1,Bạc Liêu,9,Bắc Ninh,46,Bắc Trung Bộ,7,Bài Toán Hay,5,Balkan,37,Baltic Way,30,BAMO,1,Bất Đẳng Thức,66,Bến Tre,46,Benelux,13,Bình Định,43,Bình Dương,21,Bình Phước,38,Bình Thuận,34,Birch,1,Booklet,11,Bosnia Herzegovina,3,BoxMath,3,Brazil,2,Bùi Đắc Hiên,1,Bùi Thị Thiện Mỹ,1,Bùi Văn Tuyên,1,Bùi Xuân Diệu,1,Bulgaria,5,Buôn Ma Thuột,1,BxMO,12,Cà Mau,13,Cần Thơ,14,Canada,39,Cao Bằng,6,Cao Quang Minh,1,Câu Chuyện Toán Học,36,Caucasus,2,CGMO,10,China,10,Chọn Đội Tuyển,347,Chu Tuấn Anh,1,Chuyên Đề,124,Chuyên Sư Phạm,31,Chuyên Trần Hưng Đạo,3,Collection,8,College Mathematic,1,Concours,1,Cono Sur,1,Contest,610,Correspondence,1,Cosmin Poahata,1,Crux,2,Czech-Polish-Slovak,25,Đà Nẵng,39,Đa Thức,2,Đại Số,20,Đắk Lắk,54,Đắk Nông,7,Đan Phượng,1,Danube,7,Đào Thái Hiệp,1,ĐBSCL,2,Đề Thi HSG,1637,Đề Thi JMO,1,Điện Biên,8,Định Lý,1,Định Lý Beaty,1,Đỗ Hữu Đức Thịnh,1,Do Thái,3,Doãn Quang Tiến,4,Đoàn Quỳnh,1,Đoàn Văn Trung,1,Đống Đa,4,Đồng Nai,49,Đồng Tháp,51,Du Hiền Vinh,1,Đức,1,Duyên Hải Bắc Bộ,25,E-Book,33,EGMO,16,ELMO,19,EMC,8,Epsilon,1,Estonian,5,Euler,1,Evan Chen,1,Fermat,3,Finland,4,Forum Of Geometry,2,Furstenberg,1,G. 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Mildorf,1,THPT Chuyên Lê Quý Đôn,1,THPTQG,15,THTT,6,Thừa Thiên Huế,35,Tiền Giang,19,Tin Tức Toán Học,1,Titu Andreescu,2,Toán 12,7,Toán Cao Cấp,3,Toán Chuyên,2,Toán Rời Rạc,5,Toán Tuổi Thơ,3,Tôn Ngọc Minh Quân,2,TOT,1,TPHCM,124,Trà Vinh,5,Trắc Nghiệm,1,Trắc Nghiệm Toán,2,Trại Hè,34,Trại Hè Hùng Vương,25,Trại Hè Phương Nam,5,Trần Đăng Phúc,1,Trần Minh Hiền,2,Trần Nam Dũng,9,Trần Phương,1,Trần Quang Hùng,1,Trần Quốc Anh,2,Trần Quốc Luật,1,Trần Quốc Nghĩa,1,Trần Tiến Tự,1,Trịnh Đào Chiến,2,Trung Quốc,12,Trường Đông,19,Trường Hè,7,Trường Thu,1,Trường Xuân,2,TST,55,Tuyên Quang,6,Tuyển Sinh,3,Tuyển Tập,44,Tuymaada,4,Undergraduate,66,USA,44,USAJMO,10,USATST,7,Uzbekistan,1,Vasile Cîrtoaje,4,Vật Lý,1,Viện Toán Học,2,Vietnam,4,Viktor Prasolov,1,VIMF,1,Vinh,27,Vĩnh Long,20,Vĩnh Phúc,63,Virginia Tech,1,VLTT,1,VMEO,4,VMF,12,VMO,46,VNTST,22,Võ Anh Khoa,1,Võ Quốc Bá Cẩn,26,Võ Thành Văn,1,Vojtěch Jarník,6,Vũ Hữu Bình,7,Vương Trung Dũng,1,WFNMC Journal,1,Wiles,1,Yên Bái,17,Yên Định,1,Yên Thành,1,Zhautykov,11,Zhou Yuan Zhe,1,
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MOlympiad: [Solutions] United States of America TST Selection Test 2014
[Solutions] United States of America TST Selection Test 2014
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