IMO 2014 #4

by Wolstenholme, Oct 27, 2014, 2:31 AM

Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$ . Let $M$ and $N$ be the points on $AP$ and $AQ$ , respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$ . Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$ .

Proof:

We proceed with barycentric coordinates. Let $A = (1, 0, 0)$ and $B = (0, 1, 0)$ and $C = (0, 0, 1)$ . Also let $a = BC, b = CA$ , and $c = AB$ . Since $\triangle{BAP} \sim \triangle{BCA}$ we get that $BP = \frac{c^2}{a}$ and so $P = (0 : a^2 - c^2 : c^2)$ . Similarly, $Q = (0 : a^2 - b^2 : b^2)$ . It now suffices to show that this point satisfies the equation $a^2yz + b^2zx + c^2xy = 0$ which is trivial.

This means that $M = (-a^2 : 2a^2 - 2c^2 : 2c^2)$ and $N = (-a^2 : 2a^2 - 2b^2 : 2b^2)$ . Therefore line $BM$ has equation $z = -\frac{2c^2}{a^2}x$ and line $CN$ has equation $y = -\frac{2b^2}{a^2}x$ . Therefore their intersection has coordinates $(a^2, -2b^2, -2c^2)$ .

It now suffices to show that this point satisfies the equation $a^2yz + b^2zx + c^2xy = 0$ which is trivial.

This post has been edited 1 time. Last edited by Wolstenholme, Oct 27, 2014, 2:32 AM

Comment

0 Comments

Submit

glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
the random chinese tst problem is the only thing I read, but I'll assume your blog is nice and give you a shout even though you probably never use aops anymoer
by fukano_2, Jun 14, 2020, 6:24 AM
wolstenholme - op
by AopsUser101, Jan 29, 2020, 8:27 PM
this blog is so hot
by mathleticguyyy, Jun 5, 2019, 8:26 PM
Hi. Nice Blog!
by User360702, Jan 10, 2019, 6:03 PM
helloooooo
by songssari, Jun 12, 2016, 8:21 AM
shouts make blogs happier
by briantix, Mar 18, 2016, 9:57 PM
You were just featured on AoPS's facebook page.
by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

18 shouts

Wolstenholme's blog

IMO 2014 #4

by Wolstenholme, Oct 27, 2014, 2:31 AM

0 Comments