This is the first problem of #PeruMOTraining, you can see my first post here. I proposed this problem for the Peruvian Mathematical Olympiad, in 2011. Please post your solutions!

**Problem**
Let \(ABC\) be a right triangle, with \(\angle ABC=90^\circ\). Let \(CM\) and \(AN\) be interior bisectors intersecting at \(I\) (\(M\) is on the segment \(AB\) and \(N\) is on the segment \(BC\) ). Construct the paralellograms \(AMIP\) and \(CNIQ\). If \(U\) and \(V\) are the midpoints of segments \(AC\) and \(PQ\), respectively. Prove that \(UV\) and \(AC\) are perpendicular.

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TopNewestLet's assume AB=a, BC=b and CA=c. X is the point so that VX is parallel to AB and UX is parallel to BC. So VXU is a right triangle at X. Let r be the inradius of ABC, so r = (a+b-c)/2. Let's calculate VX and UX.

VX = (1/2)IP + r - (1/2)AB = (1/2) (AM + 2r - a) = (1/2) (AM + (b-c))

By the angle bisector theorem AM/MB = c/b, so AM/a=c/(b+c), or AM = ac/(b+c) . Hence

(b+c)(AM + (b-c)) = ac + (b^2-c^2) = ac - a^2 = a(c-a) = a(c-a)(c+a)/(a+c) = ab^2/(a+c)

So 2(a+c)(b+c)VX = ab^2 Similarly 2(a+c)(b+c)UX = ba^2

So VX/UX= b/a = CB/AB. Hence VXU is similar to CBA. This should be enough to conclude UV is perpendicular to AC. – George G · 3 years, 1 month ago

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– Jorge Tipe · 3 years, 1 month ago

Please, can you explain why \(VX= (1/2)IP + r - (1/2)AB \) ?Log in to reply

The (distance from V to IQ) = (1/2)IP

The (distance from I to BC) = r

The (distance from U to BC) = (1/2)AB

VX = (distance from V to BC) - (distance from U to BC) – George G · 3 years, 1 month ago

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Wow I'm so happy that the new features have given us not only geometry problems, but also more discussions.

This is a very nice problem here I will give a synthetic solution:

Let's take advantage of all these midpoints. We construct a point on ray\(AV\) such that \(V\) is the midpoint of \(AK\).Now we have \(VU\parallel CK\), which means we just have to prove \(CK\perp AC\). Moreover, we get parallelogram \(APKQ\) so \(QK=MI, IN=QC, \angle KQC=\angle CIN=45\). Now since \(\angle QCA=\angle CAN=\angle NAB\), therefore \(CK\perp AC\) is equivalent to proving \(\angle KCQ=90-\angle QCA=90-\angle NAB=\angle ANB\). Since \(\angle KQC=45=\angle NBI\), thus we just have to prove \(\triangle KQC\sim \triangle IBN\) or simply \(\frac {BI}{BN}=\frac {QK}{QC}=\frac {MI}{IN}\). This is indeed true from \(\triangle MIB\sim \triangle INB\) and here's its proof: since \(\angle AIM=\angle ABI=45\), therefore \(\angle NIB=\angle IAM+\angle IBA=\angle IAM+\angle AIM=\angle IMB\). It's already known that \(\angle IBM=\angle IBN\), so that gives us the similarity. \(\Box\)

I haven't been keeping up with all these posts lately, wish I had found this sooner. :) – Xuming Liang · 3 years ago

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– Jorge Tipe · 3 years ago

Nice solution Xuming! You find another property of a right triangle: triangles \(MBI\) and \(IBN\) are similar!Log in to reply

– Xuming Liang · 3 years ago

I actually have a few more. I found them while trying different constructions of points and lines.Log in to reply

The following is a nice property about a right triangle.

[Keeping the same notation of our problem]

PropertyThe projections of the segments \(MI\) and \(IN\) on the line \(AC\) have the same length.

ProofLet \(E\) and \(F\) in \(AC\) such that \(ME\) and \(NF\) are perpendicular to \(AC\). Denoting \(d(X,YZ)\) the distance from point \(X\) to line \(YZ\), we have \(d(I, ME)=d(I,MB)=r=d(I, NB)=d(I,NF)\) . Since \(d(I,ME)\) is the length of the projection of \(MI\) on the line \(AC\) and \(d(I,NF)\) is the length of the projection of \(NI\) on the line (AC), we are done!Could you use this property to solve our problem?– Jorge Tipe · 3 years, 1 month agoLog in to reply

– George G · 3 years, 1 month ago

OK, so if X, Y are on AC such that both PX and QY are perpendicular to AC, what you proved above implies that AX=CY as both are projections of AP and CQ onto AC. Hence U is the midpoint of XY. Since V is the midpoint of PQ, VU is parallel to PX. Hence VU is perpendicular to AC.Log in to reply

– George G · 3 years, 1 month ago

The lemma seems to be unnecessary. Let X and Y as above, then \(\angle{PAX}=\angle{ACM} =\frac{C}{2}\). So \(AX = AP\cos{\frac{C}{2}} = IM\cos{\frac{C}{2}} =r\). Similarly \(CY=r\). The rest is the same as above.Log in to reply

– Jorge Tipe · 3 years, 1 month ago

Basically the lemma states that \(AX=CY\). My idea was to stand out a property concerning the points M, I, N without mention the other points.Log in to reply

– George G · 3 years, 1 month ago

Yes, Thank you. A proof along that line was not possible for me without seeing your lemma.Log in to reply

Are we allowed to use co-ordinate geometry? I thought of a solution using co-ordinate geometry but it is turning out to be a bit ugly. – Bruce Wayne · 3 years, 1 month ago

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– Jorge Tipe · 3 years, 1 month ago

Since we have a right angle, a solution with co-ordinate geometry seems possible.Log in to reply

– George G · 3 years, 1 month ago

Yes, I did with coordinates too, finding the coordinates of V is pretty much same as what I posted above. Then it's just a matter of verifying the slope of VU is the negative reciprocal of the slope of AC.Log in to reply

– Daniel Liu · 3 years, 1 month ago

I believe so. All solutions are welcome.Log in to reply

Please post a solution that is understandable and graceful......... I'm still in Middle School – Rohitas Bansal · 3 years, 1 month ago

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AC is parallel PQ UV &AC are perpendicular – Sathiya Narayanan · 3 years, 1 month ago

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http://i.imgur.com/1RSGjPt.png

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– George G · 3 years, 1 month ago

Even AC and PQ are parallel can't conclude that VU and AC are perpendicular unless AP=CQ which is clearly not true.Log in to reply

– Xuming Liang · 3 years ago

And AP=CQ implies that it's a 45-45-90 triangle... I'm posting this so we can have 3 people from San Diego replying to the same post. :) Such a rare scene.Log in to reply

– Andrew Tiu · 3 years, 1 month ago

Best and shortest solution here I think.Log in to reply