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

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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>.
 
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>.
  
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Part 1:
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3. As point D moves along the base of the side AB, the locus of points where C can exist fills in a semicircle because C must be a distance <math>\sqrt{(AD)(DB)} = \sqrt{(\frac{AB}{2}-k)(\frac{AB}{2}+k)} = \sqrt{(\frac{AB}{2})^2-k^2}</math>
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Part 1: If <math>\sin{A}\sin{B} \leq \sin^2{\frac{C}{2}}</math>, then such a point D exists.
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First note that in order for D to exist, the line segment CD must be able to reach from the vertex C to the side AB, meaning that

Revision as of 17:04, 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 $\frac{c}{2}$.

3. As point D moves along the base of the side AB, the locus of points where C can exist fills in a semicircle because C must be a distance $\sqrt{(AD)(DB)} = \sqrt{(\frac{AB}{2}-k)(\frac{AB}{2}+k)} = \sqrt{(\frac{AB}{2})^2-k^2}$

Part 1: If $\sin{A}\sin{B} \leq \sin^2{\frac{C}{2}}$, then such a point D exists.

First note that in order for D to exist, the line segment CD must be able to reach from the vertex C to the side AB, meaning that

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