Verified answer

Here is a geometric approach and probably the simplest method.

The trick that cracks this problem is recognising the law of cosines.

c²=a²+b²-2abcosγ

so for the first term,

c_1 = √(49 + x² - 7√(2)x) = √(7² + x² - 14xcos45)

similarly,

c_2 =√(x² + b² - √(2)xb) = √(x² + b² - 2bxcos45)

c_3 = √(50 + b² - 10b) = √((√50)² + b² - 2×√50×b×cos45)

So from the above three expressions we have three triangles with 2 that share side b and 2 that share side x (you'll have to draw them yourself) and we wish to minimise (c_1+c_2+c_3). If you look at your diagram, clearly, this value is minimised when c_1, c_2, c_3 forms a straight line which makes a larger traingle with sides √50, 7 and (c_1+c_2+c_3). If you want to prove this just use the triangle inequality. Applying the law of cosines one more time to this larger triangle:

c_1+c_2+c_3=√((√50)²+7² - 2×√50×7cos135)

=13

I'm still looking for a solution using the classical inequalities, eg. minkowski, means, and possibly geometric inequalities. To the previous answerer I tried applying AM-QM to the expression you derived but didn't work out nicely cause of equality case; I think you may still find a way to solve it though - good luck. Unfortunately, during the competition all I tried was AM-GM and Cauchy-Schwarz, got no-where, realised I'd wasted 5 min, and gave up lol. But seriously 75 minutes is crazy time constraints for doing all those questions.

Edit:

I think another approach is possible which only uses the triangle inequality and follows from the idea of the previous answerer.

Note that we can re-write

y = √(49 + x² - 7√(2)x) + √(x² + b² - √(2)xb) + √(50 + b² - 10b)

as

√((x/√2)²+(x/√2-7)²) + √((x/√2)²+(x/√2-b)²) + √((b-5)² +5²)

(Thanks to the previous answerer (who deleted answer aargh) as this expression is similar to what he derived).

By the converse of the Pythagorean Theorem, we see that c_1, c_2 and c_3 are simply the hypotenuses of 3 right triangles. We then apply the triangle inequality to determine a minimum. I have still to try this approach if I have time.

I think that there are other solutions (maybe co-ordinates, various geometry etc).

I got a*b = ab = 396 Similarity considerations allow us the dimensions on the diagram to be: |OL| = 1, |OM| = 2, |ON| = 3 It's well-known (and easy to prove) that for every internal point O the sum |OL| + |OM| + |ON| = 6 is equal to the length of the altitude of the equilateral triangle PQR, so its side length is 4√3, hence Area(PQR) = 12√3 Let's connect O with P and consider the angles λ = angle(OPL) and ν = angle(OPN), obviously (*) λ + ν = π/3 and sin λ : sin ν = |OL| : |ON| = 1 : 3 Now Area(LONP) = Area(OPL) + Area(OPN) = = |OL| * |PL| / 2 + |ON| * |PN| / 2 = = (cot λ + 9 cot ν)/2 Further sin v / sin λ = sin(π/3 - λ) / sin λ = = (√3/2) * cot λ - 1/2 = 3, so we get cot λ = 7/√3 and then, using (*), cot ν = cot(π/3 - λ) = 5√3/9 Now the irreducible fraction a / b = Area(LONP) / Area(PQR) = = (1/2)( cot λ + 9 cot ν) / (12√3) = = (11√3/3) / (12√3) = 11 / 36, so a = 11, b = 36 P.S.(After having read Philo's answer, the end of the 2nd paragraph - last 2 text lines): "√3 / 7 = sin α / cos α = tan α since 1/PL = tan α, PL = √3/7 and area POL = √3/14" Well, tan α = √3/7 indeed (λ in my notations above), but since 1/PL = tan α, I get PL = 1/ tan α = 7/√3 instead of √3/7, or I'm wrong?!? P.S.2.(Plagiarizing Lester L) If O is an internal point for the equilateral triangle PQR, then the 3 lines through O, parallel to the sides PQ, QR, RP decompose PQR into 3 parallelograms and 3 triangles - let the latter altitudes are in ratio l : m : n /l, m, n > 0/, then easily can be seen, their areas A_{l}, A_{m}, A_{n} satisfy: √A_{l} : √A_{m} : √A_{n} = l : m : n and √A_{l} + √A_{m} + √A_{n} = √Area(PQR) Hence the ratios of the roots of areas of the small triangles to the area of the big one are l/(l + m + n), m/(l + m + n) and n/(l + m + n) Then the required ratio (2 adjacent sides of the quadrilateral in ratio l : n) is (l + n)²/(l + m + n)² - (1/2)l²/(l + m + n)² - (1/2)n²/(l + m + n)² = =(2ln + l²/2 + n²/2) / (l + m + n)², the latter for l = 1, m = 2, n = 3 yields a/b = (6 + 1/2 + 9/2)/36 = 11/36, a + b =47. Lester L doesn't cease astonishing me with inventive and inspired solutions! Congratulations, Lester - reading Yours, I didn't like mine first!