Difference between revisions of "2014 USAJMO Problems"

Revision as of 23:09, 29 April 2014

Day 1

Problem 1

Let $a$ , $b$ , $c$ be real numbers greater than or equal to $1$ . Prove that $\min{\left (\frac{10a^2-5a+1}{b^2-5b+1},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc.$ Solution

Problem 2

Let $\triangle{ABC}$ be a non-equilateral, acute triangle with $\angle A=60^\circ$ , and let $O$ and $H$ denote the circumcenter and orthocenter of $\triangle{ABC}$ , respectively.

(a) Prove that line $OH$ intersects both segments $AB$ and $AC$ .

(b) Line $OH$ intersects segments $AB$ and $AC$ at $P$ and $Q$ , respectively. Denote by $s$ and $t$ the respective areas of triangle $APQ$ and quadrilateral $BPQC$ . Determine the range of possible values for $s/t$ .

Solution

Problem 3

Let $\mathbb{Z}$ be the set of integers. Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that $xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))$ for all $x, y \in \mathbb{Z}$ with $x \neq 0$ .

Solution

Day 2

@@ Line 2: / Line 2: @@
 ===Problem 1===
-Let <math>a</math>, <math>b</math>, <math>c</math> be real numbers greater than or equal to <math>1</math>. Prove that <cmath>\min{\left (\frac{10a^2-5a+1}{b^2-5b+1},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc </cmath>
+Let <math>a</math>, <math>b</math>, <math>c</math> be real numbers greater than or equal to <math>1</math>. Prove that <cmath>\min{\left (\frac{10a^2-5a+1}{b^2-5b+1},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc. </cmath>
 [[2014 USAJMO Problems/Problem 1|Solution]]

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Difference between revisions of "2014 USAJMO Problems"

Revision as of 23:09, 29 April 2014

Contents

Day 1

Problem 1

Problem 2

Problem 3

Day 2

Problem 4

Problem 5

Problem 6