MPMS Problem Solving Marathon

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Fm Jakaria
Posts: 79
Joined: Thu Feb 28, 2013 11:49 pm

Re: MPMS Problem Solving Marathon

Unread post by Fm Jakaria » Tue Jan 08, 2019 3:59 pm

Phlembac Adib Hasan wrote: Problem 5
A Hydra has 2019 heads and is immune to damage from conventional weapons. However, with one blow of a magical sword, Hercules can cut off its 9, 10, 11 or 12 heads. In each of these cases, 5, 18, 7 and 0 heads grow on its shoulder. The Hydra will die only if all the heads are cut off. Can Hercules kill the Hydra with his sword?
After each strike of Hercules, the Hydra either loses $9-5=4$ heads, or gains $18-10=8$ heads, or loses $11-7=4$ heads, or loses $12-0=12$ heads, respectively. Since the number of heads to start with, namely $2019$, is odd, we would always end up with an odd number of heads - after any number of strikes. So the count of heads can never be zero.
You made that problem up, didn't you? ;)

Problem 6
An isosceles trapezoid has equal sides of length $a$, parallel sides of lengths $b$ and $c$, and diagonals of length $d$. Without using Ptolemy's theorem, show that $d^2 = a^2 + bc$.
You cannot say if I fail to recite-
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.

aritra barua
Posts: 57
Joined: Sun Dec 11, 2016 2:01 pm

Re: MPMS Problem Solving Marathon

Unread post by aritra barua » Tue Jan 08, 2019 4:19 pm

Produce $DA$ upto $E$ such that $AE=BC$.It follows that $AEBC$ is a parallelogram.So,$BE=AC=BD=d$.By Stewart's theorem on $\bigtriangleup BED,d^2b+d^2c=m(a^2+bc)$ where $m=b+c$.Cancelling $b+c=m$ from both sides we finally have $d^2=a^2+bc$.

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