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Inside triangle $ABC$, point $O$ is selected; vectors of length $1$ are plotted on rays $OA$, $OB$, and $OC$. Prove that the length of the sum of these vectors is less than $1$.

EDIT: I know pure geometrical solution of this problem but it seems me too complex. I tried to compute a scalar square of the sum and what is needed is to prove that $\cos\alpha+\cos\beta+cos\gamma$ is less than $-1$, where $\alpha, \beta, \gamma$ are angles between the vectors.

Andrei
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eug100
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    What have you tried? Where are you stuck? Is OS the same as OC? – Andrei Jan 12 '24 at 14:15
  • It is the case $n=3$ of this very recent question which, by the way, hasn't received an answer. – Jean Marie Jan 12 '24 at 15:28
  • Let $\hat{a}, \hat{b}, \hat{c}$ be the three unit vectors. In order for $\eta(O) \stackrel{def}{=} |\hat{a} + \hat{b} + \hat{c}| = 1$, a pair of them need to be opposite to each other. This is impossible when $O$ lies in interior of triangle. Since $\eta$ depends on $O$ continuously, either all $\eta(O) < 1$ or all $\eta(O) > 1$. You just need to show there is an $O$ satisfies $\eta(O) < 1$ and you are done. – achille hui Jan 12 '24 at 15:54
  • Thank you, @Andrei, I fixed the misprint: changed S to C. I know pure geometrical solution of this problem but it seems me too complex. I tried to compute a scalar square of the sum and what is needed is to prove that cos(α)+cos(β)+cos(γ) is less that -1, where α, β, γ are angles between the vectors. – eug100 Jan 12 '24 at 15:58
  • Are you OK with a calculus based method? – Andrei Jan 12 '24 at 16:35
  • @Andrei: it might be OK. – eug100 Jan 12 '24 at 17:37
  • @eug100 I've moved your contribution from comments to the question. – Andrei Jan 12 '24 at 18:14
  • @eug100, You can find several different solutions to your problem here and here. – kabenyuk Jan 13 '24 at 09:24
  • @kabenyuk: thank you for this information. "Another proof for the 2D case" of Reza Rajaei is really impressive. – eug100 Jan 13 '24 at 16:25

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Since the point $O$ is in the interior of the triangle, it means that all angles are between $0$ and $\pi$, and the sum of the angles is $2\pi$. That means that the sum of any two angles is greater than $pi$. Then $$\cos\alpha+\cos\beta+\cos\gamma=\cos\alpha+\cos\beta+\cos(2\pi-\alpha-\beta)=\cos\alpha+\cos\beta+\cos(\alpha+\beta)$$ Now let's compute the extremum of this function of $\alpha$ and $\beta$. Taking the derivatives you get: $$-\sin\alpha-\sin(\alpha+\beta)=0\\-\sin\beta-\sin(\alpha+\beta)=0$$ From here $$\sin\alpha=\sin\beta$$ This is equivalent to $\alpha=\beta$, since $\alpha=\pi-\beta$ will not yield $\alpha+\beta>\pi$. Then $$\sin\alpha+\sin 2\alpha=0\\\sin\alpha+2\sin\alpha\cos\alpha=0$$ You can't have $\sin\alpha=0$ in the interior of the triangle, so $\cos\alpha=-1/2$, or $\alpha=\beta=2\pi/3$, for which $$3\cos\frac{2\pi}3=-1.5$$ Since we have an extremum for which the value is $-1.5$, we need to check the values on the boundary. In the limit $\gamma\to 0$, you get $\alpha+\beta\to2\pi$, but since you have both angles less than $\pi$, you get $\alpha\to\pi$ and $\beta\to \pi$. Then $$\cos\pi+\cos\pi+\cos 0=-1$$ In the other limit $\gamma\to \pi$ you get $\alpha+\beta\to \pi$, with $$\cos\alpha+\cos(\pi-\alpha)+\cos\pi=-1$$ Therefore $$-1.5\lt \cos\alpha+\cos\beta+\cos\gamma\lt -1$$

Andrei
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