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

• 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 – John Wayland Bales Jun 1 '16 at 3:48
• Thanks for the simulation. @JohnWaylandBales – zyx Jun 1 '16 at 4:05
• 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! – le duc quang Jun 1 '16 at 13:25
• There were three people having enough reputation for a close vote and not seeing the solution right away. – Christian Blatter Jun 1 '16 at 15:13
• @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?) – le duc quang Jun 1 '16 at 15:16

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.

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.

• 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. – A---B Jun 24 '16 at 17:45
• 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 ^^ – le duc quang Jun 26 '16 at 11:31
• @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.. – Mick Jun 26 '16 at 17:12
• 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 – le duc quang Jun 26 '16 at 17:23
• @leducquang Maybe your brother has the suitable-for-the-nineth-grader solution (given by his teacher, [other than the one posted]) by now. – Mick Jun 26 '16 at 17:31

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.

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.

• One more comment, I don't see why we need $OA > 2R$. – GAVD Jun 1 '16 at 15:15
• 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. – le duc quang Jun 1 '16 at 15:22
• 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 – A---B Jun 1 '16 at 16:40
• @ritwiksinha Geogebra showed neither "NM = MA" nor "NB ∥∥ MI". – Mick Jun 2 '16 at 2:24
• @ritwiksinha One of my tries is uploaded. See what can be done. – Mick Jun 24 '16 at 17:30

I'm lucky to find this beautiful solution, and it only uses 9th-grade level knowledge to solve. Here is the solution

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.

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