# Thread: Circle Geometry: Shortest Chord

1. ## Circle Geometry: Shortest Chord

AB, CD and XY are chords in a circle with centre O. XY cuts AB and CD in L and M, which are the midpoints of AB and CD. Prove that XY is greater than either AB or CD.

I hav no idea how to approach this problem

2. ## Extreme value problem...

Originally Posted by Lukybear
AB, CD and XY are chords in a circle with centre O. XY cuts AB and CD in L and M, which are the midpoints of AB and CD. Prove that XY is greater than either AB or CD.

I hav no idea how to approach this problem
Look at it as an extreme value problem like this:
Since both $\displaystyle AB$ and $\displaystyle XY$ are chords of the same circle that intersect in the midpoint $\displaystyle L$ of $\displaystyle AB$, you have that $\displaystyle XL\cdot LY = AL\cdot LB =: k$ and that $\displaystyle XY=XL+LY$.
Since $\displaystyle XL\cdot LY=k$, independently of the exact choice of $\displaystyle XY$, you'll find that the case where $\displaystyle XY=XL+LY$ attains its smallest value is exactly the case where $\displaystyle XL=LY=\sqrt{k}$. Since we know that $\displaystyle AL=LB=\sqrt{k}$, we can infer from this that $\displaystyle XY\geq AB=2\sqrt{k}$. Similarly for $\displaystyle XY\geq CD$.

3. Sorry about this, but i havent learnt half of this. Could you just explain to me why XL.LY = AL.LB =k?

4. Originally Posted by Lukybear
Sorry about this, but i havent learnt half of this. Could you just explain to me why XL.LY = AL.LB =k?
This is called the Chord Theorem (aka. Chord-Chord Power Theorem): it can be shown to hold by considering similar triangles, for example.
However, if what I wrote seems like so much gibberish to you, then, maybe, another approach is needed than what I have suggested. I just cannot think of a fundamentally different approach at the moment.

5. Perhaps a simple geometrical proof using circular properties would do? I have no idea, but if anyone could help me out on this that would be good.