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Being $A = (x_1,y_1)$, $B = (x_2,y_2)$, and $O = (0,0)$ in the Cartesian plane, demonstrate that, if $|OA| + |OB| = |AB|$, then $\overleftrightarrow{AB}$ passes through $O$.

"$|PQ|$" is the Euclidean distance from $P$ to $Q$.

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$|OA|+|OB| = \sqrt{x_1^2+y_1^2} +\sqrt{(-x_2)^2+(-y_2)^2}\ge \sqrt{(x_1+(-x_2))^2+(y_1+(-y_2))^2}= |AB|$ by Minkowski inequality. Equality occurs when $x_1,y_1$ are proportional to $x_2,y_2$. Translate this statement into proportion we have: $x_1 = k(-x_2), y_1 = k(-y_2)\implies \dfrac{y_1}{x_1} = \dfrac{y_2}{x_2} = -k\implies OA, OB$ have equal slope $k$, thus $O,A,B$ are collinear. Note that this question can be tackled in various ways such as completing squares, AM-GM inequality and even CS inequality. I tried to give it some new treat !!! by involving the Grandpa Minkowski !

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$$\sqrt{x_1^2 + y_1^2} + \sqrt{x_2^2 + y_2^2} = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$$

And square both sides, cancel out the terms: $$2\sqrt{x_1^2 + y_1^2}\sqrt{x_2^2 + y_2^2} = -2(x_1x_2 + y_1y_2)$$ Cancel out the 2, and square both sides again: $$(x_1^2+y_1^2)(x_2^2+y_2^2) = (x_1x_2 + y_1y_2)^2$$ Expand and cancel out the terms: $$x_1^2y_2^2 + x_2^2y_1^2 = 2x_1x_2y_1y_2$$ $$(x_1y_2 - x_2y_1)^2 = 0$$ Then assuming $y_1,y_2 \neq 0$ $$\frac{x_1}{y_1} = \frac{x_2}{y_2}$$ which means the slope of $OA$ and $OB$ is the same. Also $A$ and $B$ must lie on opposite quadrant because if they ere to lie on the same quadrant, then $x_1x_2 +y_1y_2$ would be positive, therefore our second equation cannot hold. Thus, $AB$ passes through $O$.

If $y_1=0$ then $x_1y_2=0$, and since $x_1$ cannot be 0 (otherwise $A$ is just $O$) then $y_2 = 0$, which means both $A$ and $B$ lie on the $x$-axis. Then the problem becomes easy.

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Consider $\triangle OAB$.

The lengths of the 3 sides are $|OA|, |OB|$ and $|AB|.$

Euclid:

The sum of the lengths of

2 sides of a triangle is greater than

the length of the 3rd side.

$\Rightarrow: $

$OAB$ is NOT a triangle .

Hence points $A,O, B$ are collinear.

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