 |
 |
Problem Posts |  |
 |
QUESTION
(Hong Kong Team Selection Test 2009) Let $a$, $b$, $c$ be the sides of a triangle. Determine all possible values of $$\frac{a^2+b^2+c^2}{ab+bc+ac}$$
(Hong Kong Team Selection Test 2009) Let $a$, $b$, $c$ be the sides of a triangle. Determine all possible values of $$\frac{a^2+b^2+c^2}{ab+bc+ac}$$
Laconic Solution Sketch
Apply Triangle Inequality, or Ravi Transformation.
Discussion
"Determine all possible values" problems are usually in one of two classes. In both cases, you just have to find these values and argue that the function takes on those values and nothing else. But how to approach this task differs across cases.
- There really is only a handful (countable number) of values. Look for some property of the givens which causes a cancellation or some special restriction. Think: magic!
- There is an interval of possible values. You will have to find bounds; corner and cage the function as if it were some animal. And don't forget to show that these bounds are the best possible, i.e. the function takes on every value within those bounds.
For a triangle, this can mean resorting to trigonometry: Note that $$\begin{eqnarray*}
\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ac} & = & \frac{\sin^{2}A+\sin^{2}B+\sin^{2}C}{\sin A\sin B+\sin B\sin C+\sin A\sin C}\\
& = & \frac{\sin^{2}A+\sin^{2}B+\sin^{2}\left(A+B\right)}{\sin A\sin B+\sin\left(A+B\right)\left(\sin A+\sin B\right)}\\
& = & \frac{\sin^{2}A+\sin^{2}B+\sin^{2}A\cos^{2}B+\sin^{2}B\cos^{2}A+2\sin A\sin B\cos A\cos B}{\sin A\sin B+\sin^{2}A\cos B+\sin A\sin B\cos A+\sin^{2}B\cos A+\sin A\sin B\cos B}
\end{eqnarray*}$$which doesn't look very appealing.
Sometimes, trigonometry makes things work (this trigonometric approach, for example, succeeds).
Sometimes, if you don't use it properly, it fails miserably . The prospect of finding a few nice values for this function is becoming dimmer.
What happened? Substituting $\sin C=\sin\left(A+B\right)$ sounded like a good idea, but it ruined the nice property that the numerator and the denominator have the same degree.
Now let's try something simple: try small numbers. If we had $a=3$, $b=4$, $c=5$, we'd get $\frac{a^2+b^2+c^2}{ab+bc+ac}=\frac{50}{47}$. Our intuition tells us this can't possibly be a constant for all triangles. It's now almost evident that this is a "determine all possible values" problem of the second type.
What now? We know using inequalities to bound the function is now inevitable. And we're talking about a triangle, so we'll probably have to invoke the Triangle Inequality. I've seen a fast way to bound it using Triangle Inequality raw, but I'll instead use a staple of triangle-based inequalities: what has been called the Ravi Substitution (Here's some context, and some pretty triangle-based inequality problems):
If $a$,$b$, and $c$ are the side lengths of a triangle, define $x=\frac{b+c-a}{2}$, $y=\frac{a+c-b}{2}$, $z=\frac{a+b-c}{2}$. These are all positive reals. Conversely, given positive reals $x$,$y$,$z$, there exists a unique triangle with side lengths $a=y+z$, $b=x+z$, $c=x+y$.Essentially, we can plug in the Ravi Substitution and forget about the triangular origin of our values altogether. The expression becomes $$\begin{eqnarray*}
A\left(x,y,z\right) & = & \frac{a^{2}+b^{2}+c^{2}}{ab+bc+ac}\\
& = & \frac{2x^{2}+2y^{2}+2z^{2}+2xy+2xz+2yz}{x^{2}+y^{2}+z^{2}+3xy+3xz+3yz}\\
& = & \frac{2\sum_{\mathrm{cyc}}x^{2}+2\sum_{\mathrm{cyc}}xy}{\sum_{\mathrm{cyc}}x^{2}+3\sum_{\mathrm{cyc}}xy}
\end{eqnarray*}$$
Note that the final two expressions are exactly the same; the latter uses a very convenient notation, cyclic summation: $\sum_{\mathrm{cyc}}x$ simply means $x+y+z$, $\sum_{\mathrm{cyc}}xy$ means $xy+yz+xz$ and so on.
Now we're in control of this beast. Rearrange so that $\sum_{\mathrm{cyc}}x^{2}$ and $\sum_{\mathrm{cyc}}xy$ are dominant: $$\frac{-3A+2}{A-2}=\frac{\sum_{\mathrm{cyc}}x^{2}}{\sum_{\mathrm{cyc}}xy} $$
The function $G\left(x,y,z\right):=\frac{\sum_{\mathrm{cyc}}x^{2}}{\sum_{\mathrm{cyc}}xy}$ is easy to play with. First, we can show $\sum_{\mathrm{cyc}}x^{2}\ge \sum_{\mathrm{cyc}}xy$ by the Rearrangement Inequality, or considering that $\frac{\sum_{\mathrm{cyc}}\left(x-y\right)^2}{2}\ge 0$. Now for an upper bound. But notice that if we set $y=z=1$, the function becomes $\frac{x^2+2}{2x+1}$. This is continuous and goes up to infinity as $x\rightarrow \infty$.
Back to $A$. From the above equation, we have $A=\frac{2G+2}{G+3}=2-\frac{4}{G+3}$. Now the set of possible values of $G$ is $\left[1,\infty\right)$, and it's evident now that the set of possible values of $A$ is the interval $\left[1,2\right)$.
Solution
Define $x=\frac{b+c-a}{2}$, $y=\frac{a+c-b}{2}$, $z=\frac{a+b-c}{2}$. We note that every set of positive reals $x$, $y$, $z$ can be derived from a certain triangle. The desired expression becomes $$\begin{eqnarray*}A\left(x,y,z\right)&=&\frac{2\sum_{\mathrm{cyc}}x^{2}+2\sum_{\mathrm{cyc}}xy}{\sum_{\mathrm{cyc}}x^{2}+3\sum_{\mathrm{cyc}}xy}\\&=&2-\frac{4}{G+3}\end{eqnarray*}$$ for $G\left(x,y,z\right):=\frac{\sum_{\mathrm{cyc}}x^{2}}{\sum_{\mathrm{cyc}}xy}$. From the Rearrangement Inequality, we have $G\ge1$ with equality when $x=y=z$. When $y=z=1$, $G=\frac{x^2+2}{2x+1}$ which (with appropriate $x>0$) can be any real number in $\left[1,\infty\right)$. It follows that $A$ can assume any value in $\left[1,2\right)$, and these are the only possible values of $A$.
No comments:
Post a Comment