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My brother asked me this problem, and he is studying ninth-grade. I can't solve it using primitive tools of pure geometry. Hope someone can give me a hint to solve it. Thanks.

Given a circle $(O, R)$ and $A$ is outside $(O)$ such that $OA > 2R$. Draw two tangents AB, AC of $(O)$. Let $I$ is midpoint of AB. Segment OI intersects with (O) at M. AM intersects with (O) at N, $N \neq M$. NI intersects with BC at Q. Prove that MQ perpendicular with OB

Here is the picture

enter image description here

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    $\begingroup$ This is not a solution, but a link to a Geogebra version of the problem. The point $B$ on the Geogebra simulation can be clicked and dragged to reposition it on the circle. [Geogebra version of the exercise][1] [1]: geogebra.org/o/TuJ5c2wB $\endgroup$ – John Wayland Bales Jun 1 '16 at 3:48
  • $\begingroup$ Thanks for the simulation. @JohnWaylandBales $\endgroup$ – zyx Jun 1 '16 at 4:05
  • $\begingroup$ Can anyone tell me why I received 3 votes to close this question? I want to know so I can improve the topic later. Thanks! $\endgroup$ – le duc quang Jun 1 '16 at 13:25
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    $\begingroup$ There were three people having enough reputation for a close vote and not seeing the solution right away. $\endgroup$ – Christian Blatter Jun 1 '16 at 15:13
  • $\begingroup$ @ChristianBlatter: yeah, I understand about that. I just wonder if I violate any rule, so they want to close this question (I can't see their reasons for closing, right?) $\endgroup$ – le duc quang Jun 1 '16 at 15:16
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This is NOT a solution. I just want to share some of my findings.

Construction: 1) Extend BO to cut the red circle at D; 2) DA cut the red circle at E and BM extended at F; 3) OE is joined.

enter image description here

By midpoint theorem, we have 1) OMI // DEFA; 2) BJ = JE; BM = MF.

  1. All angles marked with the same color are equal.

  2. OJMI is the line of centers of the 4 circles and BJHE is the common chord (excluding the green).

  3. H is the orthocenter of the isosceles triangle DBF.

  4. B, G, H, and M lie on the green circle.

One way to get the Job done is by showing that MQ is parallel to either BI or GH.

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  • $\begingroup$ I am also trying to prove $MQ$ \\ $BI$, i will upload my try later in the day. Thanks for uploading, i hope we get a good solution. $\endgroup$ – A---B Jun 24 '16 at 17:45
  • $\begingroup$ Interesting, I will take a closer look today and see if I can go on with your approach. Thanks for spending your time on this problem. It's not easy for 9th-grade, right ^^ $\endgroup$ – le duc quang Jun 26 '16 at 11:31
  • $\begingroup$ @leducquang It definitely is not a problem for a 9th-grade unless something is missing in the given. I have tried some other approaches but also not very successful.. $\endgroup$ – Mick Jun 26 '16 at 17:12
  • $\begingroup$ Yeah, I thought the same thing, but it is actually a problem in my young brother's notebook(without solution). I was very surprised about that, too $\endgroup$ – le duc quang Jun 26 '16 at 17:23
  • $\begingroup$ @leducquang Maybe your brother has the suitable-for-the-nineth-grader solution (given by his teacher, [other than the one posted]) by now. $\endgroup$ – Mick Jun 26 '16 at 17:31
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I found a solution using only Euclid geometry knowledge, but it's not in 9th-level, so I still hope to see some pure primitive solution.

enter image description here

To make the problem easier, I reverse the question to be like this: Let Q be the intersection of the line through M which is parallel with AB and BC. Then we just need to prove that N, Q and I are in a line

Let K is the intersection of BC and MN. BC is the polar line of A with (O), so we must have (MNKD) = -1. Therefore, (IN,IM,IK,ID) = -1. Let P is the intersection of MQ and NI and L is the intersection of MQ and IK. Because MP is parallel with ID, we must have L is midpoint of MP.

Using the triangle KBD, because I is midpoint of BD and MQ is parallel to BD, we have L is midpoint of MQ, too. So we conclude that P and Q coincide.

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  • $\begingroup$ One more comment, I don't see why we need $OA > 2R$. $\endgroup$ – GAVD Jun 1 '16 at 15:15
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    $\begingroup$ Yeah, I tried to figure out the necessity of $OA > 2R$, too. It's there in the problem. From what I saw in geogebra, that assumption may be not needed. $\endgroup$ – le duc quang Jun 1 '16 at 15:22
  • $\begingroup$ Just by looking i think if we join $N$ and $B$ and can prove that $NB$ $\parallel$ $MI$, then we are done. To do that we need to prove that $NM = MA$. Not 100 $\endgroup$ – A---B Jun 1 '16 at 16:40
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    $\begingroup$ @ritwiksinha Geogebra showed neither "NM = MA" nor "NB ∥∥ MI". $\endgroup$ – Mick Jun 2 '16 at 2:24
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    $\begingroup$ @ritwiksinha One of my tries is uploaded. See what can be done. $\endgroup$ – Mick Jun 24 '16 at 17:30
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I'm lucky to find this beautiful solution, and it only uses 9th-grade level knowledge to solve. Here is the solutionenter image description here

Let P and R are intersections of BC with OI and AM, correspondingly. Let S is the reflection of M through I.

Firstly, even using 9th-grade level, we can easily prove that ${RM \over RN} = {AM \over AN}$, therefore ${RM \over MA} = {RN \over AN}$. I is the same midpoint of both BA and MS, hence BSAM is a parallelogram, therefore BS is parallel to MR.

We have this equation chain:

$${PR \over PB} = {RM \over BS} = {RM \over MA} = {NR \over NA}$$

Therefore, we must have that PN is parallel to BA. Therefore, we have this equation chain:

$${PQ \over QB} = {PN \over BI} = {PN \over IA} = {MN \over MA}$$

Therefore, QM is parallel to BA. The problem is solved.

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  • $\begingroup$ @mick: let me know your idea $\endgroup$ – le duc quang Jun 30 '16 at 16:07
  • $\begingroup$ Thanx for the input. Will have a look at it some time later. $\endgroup$ – Mick Jun 30 '16 at 16:24
  • $\begingroup$ All the claimed results match the outcome drawings from Geogebra. Still, I don't think it is a problem for a 9th grader. $\endgroup$ – Mick Jul 4 '16 at 11:32

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