contest-mathgeometrytriangles
In $\Delta ABC$, $\angle ABC = 90^\circ$ , $D$ is the midpoint of line
$BC$. Point $P$ is on $AD$ line. $PM$ & $PN$ are respectively
perpendicular on $AB$ & $AC$. $PM$ = $2PN$, $AB = 5$, $BC$ = $a\sqrt{b}$,
where $a, b$ are positive integers. $a+b$ = ?
Source: Bangladesh Math Olympiad 2018 junior category.
I could not find any way to relate $PM$ = $2PN$ to this math.
Let $E$ be on $BC$ so $CE = CD$. Then $\angle DEC = 80^{\circ}$ so $CADE$ is cyclic, so we have by the power of the point with respect to $B$: $$a\cdot (a-d) = b\cdot (b-x)\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
By angle bisector theorem we have $${b-x\over x} = {a\over b}\implies b(b-x)= ax\;\;\;\;\;\;\;\;(2)$$
By (1) and (2) we have $$a(a-d)= ax \implies a-d= x$$
Since we are searching for $d+x = a= 2017$ we are done.
A quick computer check shows that $7272$, $8484$ and $9696$ have the same (maximum) number of divisors ($24$):
{1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72, 101, 202, 303, 404, 606, 808, 909, 1212, 1818, 2424, 3636, 7272},
{1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84, 101, 202, 303, 404, 606, 707, 1212, 1414, 2121, 2828, 4242, 8484},
{1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96, 101, 202, 303, 404, 606, 808, 1212, 1616, 2424, 3232, 4848, 9696}
so if I understand your question (which of these is minimum?), the answer is: $7272$.
The original question did not state that the solution must be done by hand, but in that case, @MishaLavrov's approach (below) is the best:
Recognize that $abab = ab \cdot 101$, so the factors of $abab$ are maximized when $ab$ has the most factors. It is a simple matter to find (by hand) that $72$, $84$ and $96$ have the same (maximum) number of factors.
The rest is easy.
Best Answer
$\triangle ABC \sim \triangle AMM' \sim \triangle PNM'$
$\frac{BC}{AB}=\frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=\sqrt{(2PN)^2-PN^2}=PN\sqrt3$
$\frac{NM'}{PN}=\sqrt3$
$\frac{BC}{AB}=\sqrt3$
$BC=AB\sqrt3=5\sqrt3$
$a=5,b=3,\color{blue}{a+b=8}$