ABC is a triangle with A=90 and C=30.Let M be the midpoint of BC. Let W be a circle passing through A tangent in M to BC. Let P be the circumcircle of ABC. W is intersecting AC in N and P in M. prove that MN is perpendicular to BC.
Problem
Source: ?
Tags: geometry, circumcircle, slope, power of a point, radical axis, Iran
YESMAths
21.01.2015 14:04
ATimo wrote: ABC is a triangle with A=90 and C=30.Let M be the midpoint of BC. Let W be a circle passing through A tangent in M to BC. Let P be the circumcircle of ABC. W is intersecting AC in N and P in M. prove that MN is perpendicular to BC. Let me correct your question and $\text{\LaTeX}\text{ify}$ it.
Question
$ABC$ is a triangle with $\angle BAC=90^{\circ}$ and $\angle ACB=30^{\circ}$. Let $M_1$ be the midpoint of $BC$. Let $W$ be a circle passing through $A$ tangent in $M_1$ to $BC$. Let $P$ be the circumcircle of $ABC$. $W$ is intersecting $AC$ in $N$ and $P$ in $M$. Prove that $MN$ is perpendicular to $BC$.
SolutionSolution
Let $M_1$ be the midpoint of $BC$. And let $C_1$ be the center of $W$. So, we first show that $C_1\in AC$.
Clearly $C_1M_1\perp BC$. And also, since $M_1$ is the center of $P$ as $ABC$ is a right triangle, we have $M_1A=M_1B=M_1C$.
So, $\angle M_1AB=\angle M_1BA=60^{\circ}$. And this implies that $\triangle AM_1B$ is equilateral. So, $\angle C_1M_1A=90^{\circ}-60^{\circ}=30^{\circ}=\angle C_1AM_1$.
But then, $\angle C_1AM_1+\angle M_1AB=90^{\circ}=\angle C_1AB=\angle CAB$. This shows that $\boxed{C_1\in AC}$.
Next $C_1N=C_1M_1$ and $\angle AC_1M_1=180^{\circ}-30^{\circ}-30^{\circ}=120^{\circ}=180^{\circ}-\angle NC_1M_1$, showing that $\triangle NC_1M_1$ is equilateral.
Again, $C_1M=C_1A$ showing $\angle C_1AM=\angle C_1MA$.
Now note that $AM$ is the radical axis of $P$ and $W$. So, $M_1C_1\perp AM$. Let $M_1C_1\cap AM=X$. Then,
$AC_1X=60^{\circ}\Rightarrow NC_1M=180^{\circ}-\angle MC_1X-60^{\circ}=60^{\circ}$.
But $C_1M=C_1N$. Thus $\angle MNC_1=\angle NC_1M\Rightarrow MN\parallel M_1C_1\Rightarrow \boxed{MN\perp BC}$.
This completes the proof.
achilles04
27.02.2015 18:09
Let $O_1$ be the center of $P$ and $O_2$ be that of $W$.
Here we see that $AB=BC/2=BO_1$ ( as $\angle ABC=60$ and $AC \perp AB$) so $AB$ is tangent to $W$. Moreover $ CA \perp AB $. So $O_2$ lies on $W$. This shows that $ \angle NMA =90$ as $AN$ is the diameter of $W$. Also we see that $AM$ is the radical axis of the two circles. Now if we prove that $AM \parallel BC $ then we are done. This is easy to prove. For this I used coordinate...
Let $AC$ be the $y-axis $ $AB$ be the $x-axis$ let $B=(c, 0), C=(0, b)=(0, \sqrt{3} c) $. So we find that equation of $W$ is $x^2+y^2- \frac{2\sqrt{3} c}{3}y=0 \implies S_1$ and that of $P$ is $x^2+y^2-cx- \sqrt{3} cy=0 \implies S_2$, so equation of radical axis $S_1-S_2$ we find the slope of this radical axis and see that it is equal to $- \sqrt{3}=$ slope of $BC$. This proves that $AM \parallel BC$ or $MN \perp BC$
"Iranian geometry olympiad senior level problem 1"..
Moderators could probably add the source...
utkarshgupta
02.03.2015 17:31
Solution : $\angle M_1MA = \angle ABM_1B = 60$ ($BC$ is tangent to $W$) But since $MM_1 = AM_1$ = Radius of $P$ $\implies AMM_1$ is equilateral $\implies ANM = \angle AM_1M = 60$ $\implies MN$ is perpendicular to $BC$ Simple and direct
aditya21
04.03.2015 19:23
let $D$ be intersection of $MN,BC$. first let $R$ be circumradius of triangle $ABC$ than $AB=BM_1=R$ hence $AB$ is tangent to $W$ $\angle BCA =\angle CAM_1=\angle CM_1N= 30$ $\angle BAM_1=\angle AMM_1=\angle MAM_1=\angle M_1ND=60$ thus now in triangle $M_1ND$ we easily get $\angle NDM_1=90$ and hence $MN$ is perpendicular to $BC$.
darko.milev
31.08.2018 22:06
Why please someone tell me <CAM1 = <CM1N ? In the solution from aditya21.
george_54
01.09.2018 00:25
darko.milev wrote: Why please someone tell me <CAM1 = <CM1N ? In the solution from aditya21. It' s because the median from the right angle is the half of the hypotenuse and $CN=\dfrac{AC}{3}$ (or relation between an inscribed angle and an angle made of a chord and a tangent).
Catoptrics
06.05.2020 17:31
redacted.
zuss77
12.07.2020 22:40
If $A$ - midpoint of $BD$, then $\triangle BCD$ - equilateral and $W$ - its incircle. If $O$ - center of $W$ then $O\in AC, AO=ON=NC$ and $MN\parallel DO$ as midline in $\triangle DOC$.
jchang0313
11.10.2020 12:09
Extend $MN$ to meet $BC$ at $X$. Firstly,$\triangle BAM_1$ is equilateral, and $\triangle ACM_1$ is isosceles. $\angle BM_1A=\angle AMM_1=\angle MAM_1=\angle MM_1C=60$ $\angle BCA=\angle M_1CA=\angle CAM_1=\angle M_1MN=\angle M_1MX=30$ Considering$\triangle M_1MX$, $\angle MXM_1=90$ thus $MN$ is perpendicular to $BC$.
BelieverofMaths
23.07.2021 14:44
ATimo wrote: ABC is a triangle with A=90 and C=30.Let M be the midpoint of BC. Let W be a circle passing through A tangent in M to BC. Let P be the circumcircle of ABC. W is intersecting AC in N and P in M. prove that MN is perpendicular to BC. Denote $ O $ as the midpoint of $BC $ and $O'$ to be the circle $W $ then $\angle {O`OA} = \angle {O`OM} = 30 $ because $O' $ is the centre of $W$ and is tangent to $BC$ at $O$ . Therefore $\angle {MNA} = \angle{MOA} = 60 $ . Also $\angle {MNA} = \angle {CNX} $ , where $ X$ is the intersection of $MN$ with $BC$ . Hence MN is perpendicular to BC.
AhmadGgx
26.08.2023 15:30
ATimo wrote: ABC is a triangle with A=90 and C=30.Let M be the midpoint of BC. Let W be a circle passing through A tangent in M to BC. Let P be the circumcircle of ABC. W is intersecting AC in N and P in M. prove that MN is perpendicular to BC. How is this classified as high school olympiad?