sum < MP_iA =1/2 <BAC when BP_1=P_1P_2= ... =P_ {n-1}C=1/n BC, isosceles

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sum < MP_iA =1/2 <BAC when BP_1=P_1P_2= ... =P_ {n-1}C=1/n BC, isosceles

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Source: 2010 Romania JBMO TST1 P4

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#1 h May 25, 2020, 2:56 AM

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Let $ABC$ be an isosceles triangle with $AB = AC$ and let $n$ be a natural number, $n>1$ . On the side $AB$ we consider the point $M$ such that $n \cdot AM = AB$ . On the side $BC$ we consider the points $P_1, P_2, ....., P_ {n-1}$ such that $BP_1 = P_1P_2 = .... = P_ {n-1} C = \frac{1}{n} BC$ .
Show that: $\angle {MP_1A} + \angle {MP_2A} + .... + \angle {MP_ {n-1} A} = \frac{1} {2} \angle {BAC}$ .

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#2 h May 25, 2020, 7:27 AM

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Set $A=(0,b), B= (a,0)$ and $C=(-a,0)$
Then $M = \left( \frac{a}{n} , \frac{b(n-1)}{n} \right)$
and $P_i = \left( a- \frac{2a}{n} i , 0 \right)$
${\alpha}_i =$ Slope $(MP_i) = \frac{b(n-1)}{a(2i-(n-1))}$
${\beta}_i =$ Slope $(AP_i) = \frac{bn}{a(2i-n)}$
$\therefore \angle MP_iA = \cot^{-1} \left( \frac{1}{\beta_i } \right) - \cot^{-1} \left(\frac{1}{ \alpha_i} \right)$
$\Rightarrow \angle MP_iA = \cot^{-1} \left( \frac{a(2i-n)}{bn} \right) - \cot^{-1} \left( \frac{a(2i-(n-1))}{b(n-1)} \right)$
$\therefore \sum_{i=1}^{n-1} \angle MP_iA = \sum_{i=1}^{n-1} \cot^{-1} \left( \frac{a(2i-n)}{bn} \right) - \cot^{-1} \left( \frac{a(2i-(n-1))}{b(n-1)} \right)$ Let $\frac{b}{a} = x$ we have $\frac{1}{2} \angle BAC = \frac{\pi}{2} - \cot^{-1} \left( \frac{1}{x} \right)$
So the we have to prove that
$\cot^{-1} \left( \frac{(2i-n)}{nx} \right) - \cot^{-1} \left( \frac{2i-(n-1)}{(n-1)x} \right) = \frac{\pi}{2} - \cot^{-1} \left( \frac{1}{x} \right)$ $\Leftrightarrow \sum_{i=0}^{n} \cot^{-1} \left( \frac{2i-n}{nx} \right) - \sum_{i=0}^{n-1} \cot^{-1} \left( \frac{2i-(n-1)}{(n-1)x} \right) = \frac{\pi}{2}$ Now observe that $\cot^{-1} \left( \frac{2(k-i)-k}{kx} \right) = \cot^{-1} \left( \frac{k - 2i }{kx } \right) =\pi - \cot^{-1} \left( \frac{2i-k}{kx} \right)$
Now $\lim_{i \rightarrow m} \cot^{-1} \left( \frac{2i - 2m}{2mx} \right) = \frac{\pi}{2}$
$\sum_{i=0}^{k} \cot^{-1} \left( \frac{ 2i-k }{kx} \right) = \begin{cases} \left( \frac{k+1}{2}\right) \pi & \mbox{ if } k \mbox{ is odd } \\ \left( \frac{k}{2} \right) \pi & \mbox{ if } k \mbox{ is even } \end{cases}$
Now among $n$ and $n-1$ one is odd and another one is even, WLOG $n$ be even.
$\sum_{i=0}^{n} \cot^{-1} \left( \frac{ 2i-n }{ nx } \right) - \sum_{i=0}^{n-1} \cot^{-1} \left( \frac{2i-(n-1)}{(n-1)x } \right) = \frac{\pi}{2}$
$\therefore \sum_{i=1}^{n-1} \angle MP_iA = \frac{1}{2} \angle BAC$ Hence, proved.

This post has been edited 3 times. Last edited by Omega18, May 25, 2020, 10:45 AM

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#3 h Oct 7, 2020, 4:32 PM

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Reflect $M$ over the axis of symmetry of triangle $ABC$ . We need to prove $\angle MP_1M'+\angle MP_2M'+...+\angle MP_{n-1}M'=2(\angle {MP_1A} + \angle {MP_2A} + .... + \angle {MP_ {n-1} A} )=\angle BAC$ since $\angle MP_iA=\angle M'P_{n-i}A$ . Notice that by similarity we have $\frac{AM}{MM'}=\frac{AB}{BC}$ so we get $MM'=\frac{BC}{n}=P_iP_{i+1}$ . Also notice that we have a ton of parallelograms hanging in the picture. Now it's easy to finish since $MP_{n-1}||AC$ . For the sake of simplicity I'll show on $n=6$ .

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