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# Right triangle and two parallelograms

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.

Note by Jorge Tipe
3 years, 5 months ago

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Let'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. · 3 years, 5 months ago

Please, can you explain why $$VX= (1/2)IP + r - (1/2)AB$$ ? · 3 years, 5 months ago

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) · 3 years, 5 months ago

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. :) · 3 years, 5 months ago

Nice solution Xuming! You find another property of a right triangle: triangles $$MBI$$ and $$IBN$$ are similar! · 3 years, 5 months ago

I actually have a few more. I found them while trying different constructions of points and lines. · 3 years, 5 months ago

The following is a nice property about a right triangle.

[Keeping the same notation of our problem]

Property
The projections of the segments $$MI$$ and $$IN$$ on the line $$AC$$ have the same length.

Proof Let $$E$$ 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? · 3 years, 5 months 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. · 3 years, 5 months 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. · 3 years, 5 months 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. · 3 years, 5 months ago

Yes, Thank you. A proof along that line was not possible for me without seeing your lemma. · 3 years, 5 months ago

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. · 3 years, 5 months ago

Since we have a right angle, a solution with co-ordinate geometry seems possible. · 3 years, 5 months 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. · 3 years, 5 months ago

I believe so. All solutions are welcome. · 3 years, 5 months ago

Please post a solution that is understandable and graceful......... I'm still in Middle School · 3 years, 5 months ago

AC is parallel PQ UV &AC are perpendicular · 3 years, 5 months ago

In general, $$AC$$ and $$PQ$$ are not parallel.

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

· 3 years, 5 months ago

Even AC and PQ are parallel can't conclude that VU and AC are perpendicular unless AP=CQ which is clearly not true. · 3 years, 5 months 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. · 3 years, 5 months ago