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## USAMO 1973 #4August 19, 2009

Posted by lumixedia in algebra, Problem-solving.
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7 comments

A fairly straightforward algebra problem. Could appear on a modern AMC-12, though the decoy answers would have to be carefully written.

USAMO 1973 #4. Determine all the roots, real or complex, of the system of simultaneous equations $\displaystyle x+y+z=3$ $\displaystyle x^2+y^2+z^2=3$ $\displaystyle x^3+y^3+z^3=3$

## USAMO 1973 #3August 17, 2009

Posted by lumixedia in combinatorics, Problem-solving.
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1 comment so far

USAMO 1973 #3. Three distinct vertices are chosen at random from the vertices of a given regular polygon of ${(2n+1)}$ sides. If all such choices are equally likely, what is the probability that the center of the given polygon lies in the interior of the triangle determined by the three chosen random points? (more…)

## USAMO 1973 #2August 11, 2009

Posted by lumixedia in Problem-solving.
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USAMO 1973 #2. Let ${\{X_n\}}$ and ${\{Y_n\}}$ denote two sequences of integers defined as follows: $\displaystyle X_0=1,\hspace{0.1cm}X_1=1,\hspace{0.1cm}X_{n+1}=X_n+2X_{n-1}\hspace{0.1cm}(n=1,2,3,...)$ $\displaystyle Y_0=1,\hspace{0.1cm}Y_1=7,\hspace{0.1cm}Y_{n+1}=2Y_n+3Y_{n-1}\hspace{0.1cm}(n=1,2,3,...)$

Thus, the first few terms of the sequence are: $\displaystyle X:\hspace{0.1cm}1,1,3,5,11,21,...$ $\displaystyle Y:\hspace{0.1cm}1,7,17,55,161,487,...$

Prove that, except for “1”, there is no term which occurs in both sequences. (more…)

## USAMO 1973 #1August 7, 2009

Posted by lumixedia in Problem-solving.
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2 comments

USAMO 1973 #1. Two points, ${P}$ and ${Q}$, lie in the interior of a regular tetrahedron ${ABCD}$. Prove that angle ${PAQ<60^{\circ}}$. (more…)

## USAMO 1972 #5August 4, 2009

Posted by lumixedia in Problem-solving.
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USAMO 1972 #5. A given convex pentagon ${ABCDE}$ has the property that the area of each of the five triangles ${ABC}$, ${BCD}$, ${CDE}$, ${DEA}$, ${EAB}$ is unity. Show that every non-congruent pentagon with the above property has the same area, and that, furthermore, there are an infinite number of such non-congruent pentagons. (more…)

## USAMO 1972 #4July 26, 2009

Posted by lumixedia in Problem-solving.
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USAMO 1972 #4. Let ${R}$ denote a non-negative rational number. Determine a fixed set of integers ${a}$, ${b}$, ${c}$, ${d}$, ${e}$, ${f}$ such that, for every choice of ${R}$, $\displaystyle |\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt{2}|<|R-\sqrt{2}|.$ (more…)

## USAMO 1972 #2, #3July 21, 2009

Posted by lumixedia in Problem-solving.
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3 comments

I think I might as well just start going through the USAMOs in chronological/numerical order.

USAMO 1972 #2. A given tetrahedron ${ABCD}$ is isosceles, that is ${AB=CD}$, ${AC=BD}$, ${AD=BC}$. Show that the faces of the tetrahedron are acute-angled triangles. (more…)

## USAMO 2009 #5July 19, 2009

Posted by Damien Jiang in Problem-solving, Uncategorized.
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I like Olympiad geometry. Therefore, I will give my solution to this year’s USAMO #5; I was rather happy with my solution.

5. Trapezoid ${ABCD}$, with ${\overline{AB}||\overline{CD}}$, is inscribed in circle ${\omega}$ and point ${G}$ lies inside triangle ${BCD}$. Rays ${AG}$ and ${BG}$ meet ${\omega}$ again at points ${P}$ and ${Q}$, respectively. Let the line through ${G}$ parallel to ${\overline{AB}}$ intersects ${\overline{BD}}$ and ${\overline{BC}}$ at points ${R}$ and ${S}$, respectively. Prove that quadrilateral ${PQRS}$ is cyclic if and only if ${\overline{BG}}$ bisects ${\angle CBD}$.

## USAMO 1972 #1July 18, 2009

Posted by lumixedia in Problem-solving.
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9 comments

My first post was going to be an introduction to combinatorial game theory, but putting that together would have been rather more complicated than grabbing some USAMO problem and putting up my solution, so of course I chose the path of less resistance. The intro to game theory will come eventually, but in the meantime, here’s the first USAMO problem ever:

USAMO 1972 # 1. The symbols ${(a,b,...,g)}$ and ${[a,b,...,g]}$ denote the greatest common divisor and the least common multiple, respectively, of the positive integers ${a,b,...,g}$. For example, ${(3,6,18)=3}$ and ${[6,15]=30}$. Prove that $\displaystyle \frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}.$

Here is, based on my first instinct when seeing this problem… (more…)