Triangle inscribed in a circle radius 1

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Triangle inscribed in a circle radius 1

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Source: CGMO 2004 P3

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April

#1 h Dec 28, 2008, 12:01 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $ABC$ be an obtuse inscribed in a circle of radius $1$ . Prove that $\triangle ABC$ can be covered by an isosceles right-angled triangle with hypotenuse of length $\sqrt {2} + 1$ .

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#2 h Jan 1, 2009, 9:54 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Assume that $\widehat{ACB}>90^0$ and $\widehat{CAB}\leq 45^0$ .
Contruct an $E$ -isosceles right-angled triangle $AED$ with $AD=\frac{\sqrt{2}+1}{2}AB$ and $C,E$ are same side with $AB$ .
Then $AD\leq \sqrt{2}+1$ because $AB\leq 2$
We prove that: $C$ is inside triangle $AED$ .
In fact: Call $M$ is midpoint of $AB$ and $F$ is foot of peperdicular line from $M$ to $DE$ .
Then $MF=\frac{MD}{\sqrt{2}}=\frac{AD-AM}{\sqrt{2}}=\frac{AB}{2}=AM=MB$
It mean $DE$ is tangent with $(AFB)$ .
But $C$ is inside $(AFB)$ , it mean $C$ is inside triangle $AED$ .

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