A nice Olympiad qualifying question from the Philippines

√2+√6=4 sin 75°

Honestly surprised by how relatively simple this is.

I got as far as √(8+4√3) in my head but had no clue how to resolve that expression further. The progression of second half of the video was really pleasing to watch. Thanks.

Olympiad questions are special. They never require some advanced mathematics, but they are still able to let you think in a creative way, because every problem is different.

Damn after being a subscriber for two years I finally managed to solve a problem by my self; im proud of myself

When I was in middle school, we students weren’t allowed to draw anymore lines into the drawings because it would make it easier to solve it and it wouldn’t require specific algorithms anymore but now that I know this is how math problems should be solved, I’m now going practice doing this. This channel has opened my mind in math and showed me how I should approach math. +1 subscriber

When solving the two equations at the end, it is easier to take advantage of the fact that x and y are positive integers. The equations are xy = 12 and x + y = 8 There are only 3 possibilities for the first equation, 1-12, 2-6, 3-4. The only pair that sums to 8 is 2-6. Hence, x =2 and y = 6.

Wow! Presh's accent makes it legit easier to understand. Wish he taught me maths. Didn't know Olympiad had easy questions too.

Consider the intersection pt. as 'O'. The arc BO subtends angle 30° (sin(1/2)) at the centre of the left circle. Therefore, angle BAO=15° Now, AD.sin (15°)= 1 AD= √6 + √2

Few suggestions for simplification : 1. By symmetry it is quite obvious that both the line joining both centers, and AD are going through the tangent point which is center of symmetry ; from there we have 2 right triangles to calculate. 2. Once you reach xy=12, there are not so many possibilities with x,y integers - checking for x=1,2,3 is sufficient to find the single solution that also satisfies x+y=8 where x

I solved it as follows: I drew a line from D to the intersection point (called M), then the perpendicular from M to the lower line and called their intersection point K. Then I considered the two right triangles CMK and DMK, and exploited two facts: 1. CK + DK = CD = 2; 2. DK : KM = KM : CK (by the proportion of right triangles with one side in common). From 1) and 2) one can find CK, DK and, by Pythagora's theorem, DM. The result we're looking for is 2DM - the double root trick will give the desired form. The solution as Presh presented is more elegant, but this is another way nonetheless!

I love the way Presh explains things, especially when he avoids those "weird" names for standard formulae. Great video.

In the process of factoring x^2 - 8x + 12, you had to answer the question "what are two numbers such that their sum is 8 and their product is 12", which is just the question you started with in "solve 8=x+y and 12=xy". Rewriting it as a quadratic does allow you to solve using the quadratic formula, in case you can't find it in your head.

5:45 it’s a small refinement and saving, but we’re already done right there. The system of equations is symmetrical to swapping x and y, so the quadratic will necessarily have as solution the values of x and y.

I solved many problems of your channel. But I still feel so proud. Thank you.

In the last part, there is an easier way. (x-y)^2 = (x+y)^2 - 4xy so that, you just need to solve simultaneous equations x+y= 8, x-y=4

The process of solving the equation is really great. I like it very much.And also learnt a new process of equation solving without calculator.

"...our favorite right-triangle theorem..." So no specific attribution as opposed to "not Pythagoras." I guess that's progress of a sort :-)

Very nice problem .Explained beautifully .Thanks ,dear professor

Having established the length of the first triangle’s long leg (square root of 3 or 1,7320) we have for the second triangle: Short leg = 1 Long leg = 1 + 1,7320 + 1 = 3,7320 Therefore the distance A to D (the hypotenuse of this right triangle) = 3,86 Simpler