Geometry warmup: internally tangent circles

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Geometry warmup: internally tangent circles

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Source: HMMT Invitational Contest 2016, Problem 2

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6876 posts

#1 h Apr 22, 2016, 2:49 PM • 2 Y

Y by 62861, Adventure10

Let $ABC$ be an acute triangle with circumcenter $O$ , orthocenter $H$ , and circumcircle $\Omega$ . Let $M$ be the midpoint of $AH$ and $N$ the midpoint of $BH$ . Assume the points $M$ , $N$ , $O$ , $H$ are distinct and lie on a circle $\omega$ . Prove that the circles $\omega$ and $\Omega$ are internally tangent to each other.

Dhroova Aiylam and Evan Chen

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62861

#2 h Apr 22, 2016, 3:19 PM • 4 Y

Y by yojan_sushi, Arc_archer, Adventure10, Mango247

Solution

Unfortunately

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6012 posts

MSTang

#3 h Apr 22, 2016, 3:30 PM • 2 Y

Y by Adventure10, Mango247

Curious -- what sort of "joint proposal" happened for this problem?

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#4 h Apr 22, 2016, 3:34 PM • 2 Y

Y by Adventure10, Mango247

My solution,
Let $O', H'$ be the reflections of $O, H$ in $AB$ , resp. Easy to get $O'$ be center of $\odot (ABH)$ , $H'\in \Omega$ .
We have $O'M\perp AH$ , $O'N\perp BH$ $\Rightarrow HO'$ be the deameter of $\omega$ $\Rightarrow I$ (midpoint of $O'H$ ) is center of $\omega$ .
Because, $O\in \omega$ $\Rightarrow \angle OO'H=90^{\circ}$ $\Rightarrow OHH'O'$ is rectangle $\Rightarrow H'\in \omega$ and $I\in OH'$ $\Rightarrow \omega$ and $\Omega$ are tangent to at $H'$ .

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#5 h Apr 22, 2016, 4:11 PM • 6 Y

Y by MSTang, 62861, Imayormaynotknowcalculus, Adventure10, Mango247, Aryan-23

MSTang wrote:

Curious -- what sort of "joint proposal" happened for this problem?

Mostly by Dhroova. It was a problem in the HMMT database which was a few years old, and I edited it for this contest.

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Hydrogen-Helium

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Hydrogen-Helium

#6 h Apr 23, 2016, 12:49 PM • 1 Y

Y by Adventure10

This problem reminds me of the 4th question of Amir's 150 geometry problems http://cdn.artofproblemsolving.com/aops20/attachments/67223_4872d512bdc737f3379ad265e17340d1

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#7 h Apr 18, 2017, 9:38 PM • 1 Y

Y by Adventure10

Let $O'$ be the circumcenter of $\triangle HAB$ .Obviously $O'\in \odot HMN$ $\underbrace{\implies}_{OO'\perp AB}$ $\angle OMN=90^{\circ}-\angle O'NM=\angle HNM$ $\implies$ $OH||MN$ .Let $AH\cap \odot ABC={A'}$ $\implies$ $d(O,AB)=d(A',AB)=d(H,AB)$ $\implies$ $OHO'H'$ is a rectangle $\implies$ $A'\in \odot HMN$ and now homothety $\mathcal{H}_{A',2}$ ends it.

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#8 h Mar 24, 2018, 8:27 AM • 1 Y

Y by Adventure10

Interesting complex approach:

Let $L$ be a point on $\omega$ . Then if we make a homothety $h$ with center $H$ and coefficient $2,$ it sends $\omega$ to $\gamma-$ the reflection of $\Omega$ with respect to $AB$ . So the reflection with respect to $AB$ of $h(l)$ is on $\Omega.$ But the reflection of $h(l)$ is $a+b-ab(\overline{2l-h})$ . So $L$ is on $\omega$ iff $| a+b+ab\overline{h}-2ab\overline{l}|=1.$

But $O$ is on $\omega,$ so $| a+b+ab\overline{h}|=1.$ And $T=t$ is on both $\omega$ and $\Omega$ iff $|t|=1$ and $| a+b+ab\overline{h}-2ab\overline{t}|=1.$

Triangle inequality gives us:
$|2ab\overline{t}-(a+b+ab\overline{h})|+|a+b+ab\overline{h}|\ge |2ab\overline{t}|.$ Since we have the equality case, $\frac{a+b+ab\overline{h}}{2ab\overline{t}}=\frac{\frac{a+b+ab\overline{h}}{2ab}}{\overline{t}}$ is a positive integer. But $T$ is on the unit circle, so there is exactly one such point. This means that $\omega$ and $\Omega$ are tangent, and thus internally tangent (the point $H$ from $\omega$ that is in the interior of $\Omega$ ).

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pad

#9 h Dec 21, 2019, 7:09 AM • 1 Y

Y by Adventure10

We want to show that the radius of $(HMN)$ is $R/2$ since the circle contains $O$ . A homothety at $H$ with scale factor 2 sends $M\to B$ and $N\to C$ , so the radius of $(HMN)$ is half the radius of $(HBC)$ . Since $(HBC)$ is the reflection of $(ABC)$ over $BC$ , they have the same radius, done.

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#10 h Dec 21, 2019, 9:26 AM • 3 Y

Y by Pakistan, strawberry_circle, Adventure10

Notice that Circumradius of $\odot(AHB)=\frac{AB}{\sin(180^\circ-\angle AHB)}=\frac{AB}{\sin\angle ACB}=$ Circumradius of $\triangle ABC=R$ . Now notice the Homothety $\mathcal H$ cenered at $H$ mapping $M$ to $A$ and $N$ to $B$ . Hence, Radius of $\odot(AMN)=\frac{1}{2}R$ .

Now we just have to prove that any circle $(\omega_1$ ) passing through the center of a circle ( $\omega_2$ ) with radius half of $\omega_2$ , must be internally tangent to the circle $\omega_2$ .

This is trivial but to prove rigorously we will use contradiction.

Let $\omega_1\cap\omega_2=\{P,Q\}$ and $O_1,O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively. and let $O_2O_1\cap\omega_1=R$ and $O_1O_2\cap\omega_2=R'$ . Notice that $O_1R=O_1R'=\frac{R}{2}\implies R'\equiv R$ . Hence, $\omega_1$ and $\omega_2$ must be internally tangent. So we are done. $\blacksquare$

This post has been edited 3 times. Last edited by amar_04, Dec 21, 2019, 9:27 AM

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#11 h Apr 10, 2020, 10:09 PM • 1 Y

Y by Mango247

Let $R$ be the circumradius of $\triangle{ABC}$ , and let $H'\in (ABC)$ be the reflection of $H$ over $BC$ . Note that $BH'C$ has circumradius $R$ , so $\triangle{BHC}\cong \triangle{BH'C}$ has circumradius $R$ as well, and thus $\triangle{MHN}$ has circumradius $R/2$ . Hence, $\omega$ passes through $O$ and has radius $R/2$ , so we are immediately done.

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Grizzy

#12 h Oct 3, 2020, 5:37 PM

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Solution

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#13 h Mar 11, 2022, 5:32 AM

Y by

This is cool

edit: no

Notice that the nine-point circle of $\triangle ABC$ passes through $M$ , $N$ , and the reflection of $H$ over $\overline{MN}$ . Then, it must also pass through the reflection of $O$ over $\overline{MN}$ . Thus, $\omega$ has the same radius as the nine-point circle, or half the radius of $\Omega$ . Since $\omega$ passes through $O$ , $\omega$ and $\Omega$ must be internally tangent, as desired.

Remark: It seems like nobody else has put this approach on the thread yet. I thought the nine-point circle construction was quite natural because it involved $M$ and $N$ and the nine-point circle also has radius $\frac{R}{2}$ .

This post has been edited 5 times. Last edited by CyclicISLscelesTrapezoid, Nov 27, 2022, 11:02 PM

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#14 h Apr 4, 2022, 12:36 PM

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lol
It is known that the radii of $(BHC)$ and $(ABC)$ are equal, so the diameter of $(MHN)$ is equal to the radius of $(ABC)$ (since a homothety of scale factor $2$ at $H$ sends $\overline{MN}$ to $\overline{BC}$ ). But it is evident that any circle whose diameter is equal to the radius of $(ABC)$ passing through $O$ is internally tangent to $(ABC)$ , so we're done. $\blacksquare$

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#15 h Jan 18, 2023, 10:14 AM • 2 Y

Y by GeoKing, HoripodoKrishno

woah! I actually liked this problem

.

https://i.imgur.com/qs4zLDa.png

Let $N_9$ denote the $9-$ point center. So we get $N_9M=N_9N=\dfrac{R}{2}$ and thus $N_9$ lies on the perpendicular bisector of $MN$ . Also, it is well known that $N_9$ is the midpoint of $OH$ and thus again, $N_9$ lies on the perpendicular bisector of $OH$ . But as we are told $MNOH$ is cyclic, thus the intersections of the perpendicular bisectors of $\{MN,OH\}$ is the center of $\odot(MNOH)$ which is our $N_9$ .

Thus we get that $N_9O=N_9H=N_9N=N_9M=\dfrac{R}{2}$ . Moreover as $\overline{N_9-O-H}$ are collinear, we have $OH=N_9O+N_9H=\dfrac{R}{2}+\dfrac{R}{2}=R$ and thus the orthocenter $H$ lies on $\odot(ABC)$ .

Thus, one of the angles is $90^{\circ}$ , also note that $\odot(MNOH)$ becomes the $9-$ point circle of $\triangle ABC$ and $OH$ becomes the diameter of it. Thus we just have to show that in a triangle $ABC$ with one of the angles $=90^{\circ}$ , the $9-$ point circle is tangent to $\odot(ABC)$ .

WLOG $A=90^{\circ}$ . Then $A$ becomes the orthocenter of $\triangle ABC$ . Now as radius of $\odot(ABC)$ is $R$ and diameter of $MNOH$ is $R$ , its easy to see that our conclusion follows.

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#16 h Jan 30, 2023, 3:47 AM

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$(HMN)$ has half the radius of $(HBC)$ , or half the radius of $(ABC)$ . Combined with the fact that it passes through $O$ , this solves the problem.

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bjump

#17 h Aug 29, 2023, 6:00 PM

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$\frac{BC}{\sin(\angle BHC)}=\frac{BC}{\sin(180-\angle A)}=\frac{BC}{\sin(\angle A) }$ So the circumradius of $(BHC)$ is the same as $(ABC)$ which is twice of $(HMN)$ which finishes the problem.

This post has been edited 1 time. Last edited by bjump, Aug 31, 2023, 4:14 PM
Reason: Unnecessary pluralization

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#18 h Oct 2, 2023, 2:03 AM

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We can see that since the circle contains $O$ we simply need the radius to be $\frac{R}{2}$ . This means that the circumcircle of $BHC$ must have radius $R$ . Let the reflection of the orthocenter over $BC$ be $D$ . We know that $D$ lies on $\Omega$ . Thus since $\triangle BDC = \triangle BHC$ . We can see that $\Omega$ has the same radius as the circumcircle of $\triangle HBC$ . We can obviously see that since $\triangle HMN$ is homothetic to $\triangle HBC$ with ratio $1/2$ , the result follows. $\blacksquare$

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#19 h Oct 28, 2023, 5:34 AM

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Let the reflection of $H$ across line $BC$ be $H'$ and let the radius of $\Omega$ equal $R$ . It is clear that $H'$ lies on $\Omega$ and $(BH'C)$ has radius $R$ . Consequently, $(BHC)$ also has radius $R$ , so $(MHN)$ has radius $\frac{R}{2}$ . This implies the desired result. $\square$

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#20 h Dec 25, 2023, 11:14 PM • 1 Y

Y by centslordm

Notice the radius of $\omega$ is half of the radius of $(HBC)$ and therefore half the radius of $\Omega$ . Since $\omega$ also contains $O$ , it is internally tangent to $\Omega$ . $\square$

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#21 h Mar 18, 2024, 12:38 PM

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The OTIS wording has $M$ and $N$ the midpoints of $BH$ and $CH$ instead.

The radius of $(MHN)$ is half of that of $(BHC)$ by a homothety centered at $H$ . But the radius of $(BHC)$ is equal to that of $(ABC)$ . In addition, $(MHN)$ contains $O$ , so $\omega$ is internally tangent to $\Omega$ .

This post has been edited 5 times. Last edited by blueprimes, Mar 18, 2024, 12:47 PM

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#22 h Jun 9, 2024, 8:46 PM

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Note that the radius of $\triangle BHC$ is the same as $\triangle ABC$ since $\frac{\sin(\angle BHC)}{BC} = \frac{\sin{\angle A}}{BC}$ , so the radius of $(HMN)$ is $\frac{R}{2}$ . Then if a circle passes through $O$ and has radius $\frac{R}{2}$ then it must be tangent to $(ABC)$ since the only point that is $R$ away from $O$ on $(HMN)$ is the antipode of $O$ wrt $(HMN)$ , so the two circles intersect once as desired.

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#23 h Jan 8, 2025, 12:48 AM

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I am solving the OTIS version where $M, N$ are the midpoints of $BH$ and $CH.$

It is well-known that $(BHC)$ has the same radius as $(ABC),$ let this value be $R.$ Thus taking a homothety centered at $H$ with ratio $1/2$ tells us that $(MHN)$ has diameter $R,$ but since the circle also passes through $O$ we are done. QED

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eg4334

#24 h Apr 8, 2025, 3:44 AM

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By a homothety centered at $H$ and a reflection over $BC$ , we have the radius of the circle to be half that of $(ABC)$ Therefore they are internally tangent.

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#25 h Today at 7:24 AM

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Let $D$ be the foot of $A$ to $BC.$ Let $E$ be the reflection of $H$ across $D.$ By $-\frac12$ homothety about $E$ , we see that $MHNE$ is cyclic, with half the radius as the circumcircle. However, since $O$ also lies on $MHN$ , this implies tangency

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