Talk:1960 IMO Problems/Problem 3

 Let ∠ACB be x, and ∠ABC be 90-x. Let M be the midpoint on hypotenuse BC and Q and P be points such that PQ contains BC, and Q is closer to C while P is closer to B. The midpoint will always be in the middle of line QP, unless n is even or infinite, which it is not. Given such a triangle, we can express the altitude to the hypotenuse as a*cos(x)*sin(x). So, h=a*cos(x)*sin(x) We shall now denote line AM by f, where AM is the median to the hypotenuse. This means that line AM= line BM = line CM, and as BM is a/2, f=a/2. ∠MAB is 90-x while ∠MAC is x. This means that ∠AMB=2x and ∠AMC is 180-2x. The length of QP is a/n. Let ∠QAM be k and ∠PAM be z, such that ∠QAP or alpha equals k+z. This means that ∠AQM is 2x-k while ∠APM is 180-2x-z. As M is sin the middle of QP, QM=PM=a/2n. Applying the sine law on triangle AQM, we will find that: 1. sin(k)/(a/2n) = sin(2x-k)/(a/2) 2. 2n*sin(k)/a = 2*sin(2x-k)/a 3. n*sin(k) = sin(2x-k) 4. n*sin(k) = sin(2x)*cos(k)-sin(k)cos(2x) {by sin(a-b) formula} as sin(2x)=2*sin(x)*cos(x), and as h=a*cos(x)*sin(x), 2*sin(x)*cos(x)=2h/a. It follows that cos(2x)=(√(a^2 - 4h^2))/a. 5. n*sin(k)+sin(k)*(√(a^2 - 4h^2))/a)=cos(k)*(2h/a) 6. sin(k)*[n+(√(a^2 - 4h^2))/a)]=cos(k)*(2h/a) 7.tan(k)=2h/(an+√(a^2 - 4h^2)) After doing similarly with tan(z), we can use the tan(z+k) formula to find the value of tan(alpha), as alpha=z+k.