This note is related to the previous note, please have a look if you haven't read that one.
This note deals with and that will be needed in bash solutions.
Also, 2 examples of ComplexBash solutions, which might surprise you by their shortness over Pure geometry solutions.
The name of the note is , because complex numbers can be treated like vectors and that makes things easier at times. In fact, solutions that use vectors are many times very similar, except for the calculations.
Before that, I want to discuss some things about how they mark the bash solutions in Olympiads (at least in India). Take these as your rules for bashing:-
If you are stumped in a problem and are finding no way to get a solution by pure geometry, then only use bash.
If you are getting a strong feeling that the problem is not easy by pure geometry and you’ll not get going with pure geometry, start bashing.
Once you’ve started a bash solution, COMPLETE IT! Because in the marking scheme, they have one thing in mind, a bash solution is either a full mark solution or a 0.
When you are bashing, do not leave loose ends, like “The above equation reduces to zero after calculation, hence…”
You have to actually do ALL the calculations that are necessary and show the result, not just state it.
(skip if you know already)
Every complex number will have a magnitude and an angle that it makes with the real axis, called it’s argument , denoted by and always This is why we can treat complex numbers as vectors whenever required.
If , then and .
A complex number of unit modulus making an angle with real axis will be given by
Argument of all positive real numbers is and that of all negative real numbers is
Argument of all numbers of the form is for and for .
Every complex number is equivalent to a vector from Origin (complex number 0) to the point where it is denoted. Thus , a vector segment from to .
The complex number equivalent to the vector segment between and will be , and .
, adjust the addition/subtraction to bring it in the range(]
, where is the angle to be traced from in anticlockwise direction, in order to reach the direction same as . Thus in other words, .
As in above thing will be some real values, we can say for all with .
(Yeah, division by zero is not allowed in complex numbers too!)
will be a real number which will help us adjust the magnitude of our complex number.
With this much machinery, we are equipped enough to obtain some results we’ll be actually needing in geometry problems, so let’s begin the actual bashing!
The notations will be standard, (G,H,O etc) except for , which won’t be sides of triangles but the complex coordinates of the vertices respectively.
Here are that will prove helpful,
(not dealing with their proofs, we want the ‘Use’ of them more than ‘Proof’)
. Midpoint of segment between will be
. Centroid of a triangle with vertices will be .
. In a , points are circumcenter, centroid, orthocenter respectively.
As we know by Euler line, . If we take the complex numbers assigned to as , then
Now if we take the circumcircle of to be centered at origin, we get
In general, , where is circumcenter coordinates.
. If is equilateral, we have .
Also, if the names of vertices are in anticlockwise order, then is obtained by rotating by in anticlockwise direction,
. If , then you can say , where ,
Now something interesting,
Problems in this note are bit basic, but I feel you might feel them cool as the complex bashes are shorter than pure geometry... from next one there will be problems from Olympiads !
Given any , equilateral triangles are constructed externally on the sides of the triangle, to obtain points . Prove that centroid pf is same as centroid of .
– Let original triangle have complex coordinates .
Now let’s use the simple fact that is obtained by rotating by .
Now centroid of
As centroid of is also , we have proved the required result.
This will have solutions by pure geometry, but you can see they need more brain and are longer than what we did above….. But what’s more noteworthy is, imagine doing this by coordinate bash… You’d sink in the calculations to get points P,Q,R only!
On the sides of , squares are constructed externally (). Prove that centroids of triangles are all the same point.
– We use that is rotated by in anticlockwise direction in order to get .
Now observe that is just shift of by complex number
Similarly, we get all the required points, which are
Now it’s just easy calculation to find,
It’s noteworthy again that this solution is shorter and less brainy than Pure geometry and also, uses really very less calculations than Coordinate geometry…
With this, I’m concluding this part 2.
From next note onwards, there’ll be useful results and problems from past Olympiads.
If you enjoyed this, please share with your friends, who might be interested and seek help.
Stay tuned for more fun,
Happy Problem solving!