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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## Comments

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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.

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

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The (distance from V to BC) is the (distance from V to IQ) + the (distance from I to BC)

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)

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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. :)

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

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

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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?Log in to reply

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.

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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.

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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.

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

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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.

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I believe so. All solutions are welcome.

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Please post a solution that is understandable and graceful......... I'm still in Middle School

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AC is parallel PQ UV &AC are perpendicular

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In general, \(AC\) and \(PQ\) are not parallel.

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

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

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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.

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Best and shortest solution here I think.

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