Difference between revisions of "1974 IMO Problems/Problem 2"

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Since this is an "if and only if" statement, we will prove it in two parts.
 
Since this is an "if and only if" statement, we will prove it in two parts.
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Before we begin, note a few basic but important facts.
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1.  When two variables <math>x</math> and <math>y</math> are restricted by an equation <math>x+y=k</math> for some constant <math>k</math>, the maximum of their product occurs when <math>x=y=\frac{k}{2}</math>.
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2.  The triangle inequality states that for a triangle with sides <math>a</math>, <math>b</math>, and <math>c</math> fulfills <math>a + b > c</math>, meaning that <math>a > \frac{c}{2}</math> or <math>b > \frac{c}{2}</math>, which is equivalent to saying that <math>a</math> and <math>b</math> both cannot be less than or equal to <math>\frac{c}{2}</math>.
  
 
Part 1:
 
Part 1:

Revision as of 16:47, 6 January 2019

In the triangle ABC; prove that there is a point D on side AB such that CD is the geometric mean of AD and DB if and only if $\sin{A}\sin{B} \leq  \sin^2 (\frac{C}{2})$.

Solution

Since this is an "if and only if" statement, we will prove it in two parts.

Before we begin, note a few basic but important facts.

1. When two variables $x$ and $y$ are restricted by an equation $x+y=k$ for some constant $k$, the maximum of their product occurs when $x=y=\frac{k}{2}$.

2. The triangle inequality states that for a triangle with sides $a$, $b$, and $c$ fulfills $a + b > c$, meaning that $a > \frac{c}{2}$ or $b > \frac{c}{2}$, which is equivalent to saying that $a$ and $b$ both cannot be less than or equal to $\frac{c}{2}$.

Part 1: