Mnp right triangle

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Mnp right triangle

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Source: 14-th Iranian Mathematical Olympiad 1996/1997

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786 posts

Amir.S

#1 h Oct 17, 2005, 2:58 PM • 2 Y

Y by Adventure10, Mango247

Let $N$ be the midpoint of side $BC$ of triangle $ABC$ . Right isosceles triangles $ABM$ and $ACP$ are constructed outside the triangle, with bases $AB$ and $AC$ . Prove that $\triangle MNP$ is also a right isosceles triangle.

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campos

#2 h Oct 18, 2005, 4:33 AM • 1 Y

Y by Adventure10

again a trig approach, but this is the only way i can solve these problems...

let
$BC=a=2CN=2NB, AC=b=PC\sqrt2=PA\sqrt2, AB=c=MB\sqrt2=AM\sqrt2, PN=x, MN=y, PM=z, \angle BAC=\alpha, \angle ABC=\beta, \angle BCA=\theta$
now, let's apply the cosine law for $\triangle PNC$ , so we have that ${x^2=\frac{a^2}{4}+\frac{b^2}{2}-\frac{ab\cos{(\theta+45)}}{\sqrt2}=\frac{a^2}{4}+\frac{b^2}{2}-\frac{ab}{2}(\frac{a^2+b^2-c^2}{2ab}-\sin\theta})=\frac{b^2}{4}+\frac{c^2}{4}+(ABC)$
A similar procedure gives us that $y^2=\frac{b^2}{4}+\frac{c^2}{4}+(ABC)$ , so we've proved that it's isosceles.
now we have that $x^2+y^2=\frac{b^2}{2}+\frac{c^2}{2}+2(ABC)$ .
again from the cosine law applied to $\triangle APM$ we have that $z^2=\frac{b^2}{2}+\frac{c^2}{2}-bc{\cos{(\alpha+90)}}=\frac{b^2}{2}+\frac{c^2}{2}+bc\sin\alpha=\frac{b^2}{2}+\frac{c^2}{2}+2(ABC)$ , so we have that $x^2+y^2=z^2$ and we conclude that it's an right isosceles triangle.

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#3 h Oct 18, 2005, 4:23 PM • 1 Y

Y by Adventure10

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=38015 .

darij

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#4 h Oct 23, 2005, 7:59 AM • 2 Y

Y by Adventure10, Mango247

See http://www.mathlinks.ro/Forum/viewtopic.php?t=45678 Here is the particular case

$m(\angle AMB)+m(\angle APC)+m(\angle BNC)=90+90+180=360,\ i.e.$
$m(\angle MAB)+m(\angle PAC)+m(\angle NBC)=45+45+0=90.$

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#5 h Dec 1, 2021, 6:59 AM

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Let ABST and ACQR be squares outside of triangle ABC. AS and BT intersect at M. AQ and CR intersect at P. TA = BA , ∠TAC = 90+∠A = ∠BAR , AC = AR so ATC and ABR are congruent. PN = RB/2 = TC/2 = MN.

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