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Difference between revisions of "1975 USAMO Problems"

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==Problem 2==
 
==Problem 2==
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Let <math>A,B,C,D</math> denote four points in space and <math>AB</math> the distance between <math>A</math> and <math>B</math>, and so on. Show that <center><math>AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2</math>.</center>
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Let <math>A,B,C,D</math> denote four points in space and <math>AB</math> the distance between <math>A</math> and <math>B</math>, and so on. Show that
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<cmath>AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2.</cmath>
  
 
[[1975 USAMO Problems/Problem 2 | Solution]]
 
[[1975 USAMO Problems/Problem 2 | Solution]]
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== See Also ==
 
== See Also ==
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*[[USAMO Problems and Solutions]]
 
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{{USAMO box|year=1975|before=[[1974 USAMO]]|after=[[1976 USAMO]]}}
 
{{USAMO box|year=1975|before=[[1974 USAMO]]|after=[[1976 USAMO]]}}
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{{MAA Notice}}

Latest revision as of 17:57, 3 July 2013

Problems from the 1975 USAMO.

Problem 1

(a) Prove that

$[5x]+[5y]\ge [3x+y]+[3y+x]$,

where $x,y\ge 0$ and $[u]$ denotes the greatest integer $\le u$ (e.g., $[\sqrt{2}]=1$).

(b) Using (a) or otherwise, prove that

$\frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}$

is integral for all positive integral $m$ and $n$.

Solution

Problem 2

Let $A,B,C,D$ denote four points in space and $AB$ the distance between $A$ and $B$, and so on. Show that \[AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2.\]

Solution

Problem 3

If $P(x)$ denotes a polynomial of degree $n$ such that $P(k)=k/(k+1)$ for $k=0,1,2,\ldots,n$, determine $P(n+1)$.

Solution

Problem 4

Two given circles intersect in two points $P$ and $Q$. Show how to construct a segment $AB$ passing through $P$ and terminating on the two circles such that $AP\cdot PB$ is a maximum.

[asy] size(150); defaultpen(fontsize(7)); pair A=(0,0), B=(10,0), P=(4,0), Q=(3.7,-2.5); draw(A--B); draw(circumcircle(A,P,Q)); draw(circumcircle(B,P,Q)); label("A",A,(-1,1));label("P",P,(0,1.5));label("B",B,(1,1));label("Q",Q,(-0.5,-1.5)); [/asy]

Solution

Problem 5

A deck of $n$ playing cards, which contains three aces, is shuffled at random (it is assumed that all possible card distributions are equally likely). The cards are then turned up one by one from the top until the second ace appears. Prove that the expected (average) number of cards to be turned up is $(n+1)/2$.

Solution

See Also

1975 USAMO (Problems • Resources)
Preceded by
1974 USAMO 		Followed by
1976 USAMO
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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