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We got triangle $\triangle \text{ABC}$ and $\angle \text{C}=90^{\circ }$. The area of $\triangle \text{ABC}=\text{S}$ and $\angle \text{BAC}=\alpha$. Prove that the distance from the centroid to the hypotenuse is equal to $\frac{1}{2}\sqrt{\text{S}\sin\left(2\alpha \right)}$.

Any ideas how to solve that? Would be grateful if someone helps me.

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Remember the theorems "In a right triangle, the length of the median to the hypotenuse equals half the hypotenuse", and also "The centroid in a triangle is a point which divides each median in a proportion 2:1, where the longest side is always between the centroid and its vertex.

So we have

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Call the centroid $\;M\;$ and let $\;K\;$ be the midpoint of $\;AB\;$, and let $\;x=|AB|=$ the hypotenuse's length . Thus, we have:

$$\cos\alpha=\frac{AC}x\implies AC=x\cos\alpha\implies S=\frac{AC\cdot AB\cdot\sin\alpha}2=\frac{x^2\cos\alpha\sin\alpha}2=\frac14x^2\sin2\alpha$$

Now, observe that summing angles of triangle $\;\Delta BCK\;$ , we have that $$\;\angle BKC=2\alpha\implies\angle MKA=180^\circ-2\alpha\;$$

Also

$$|CK|=\frac12x\implies |MK|=\frac13|CK|=\frac x6$$

so we can know form a little right triangle $\;\Delta MKL\;$ , with $\;L\;$ a point on the hypotenuse $\;AB\;$ s.t. $\;ML\perp AB\;$ , and then in this right triangle:

$$\sin(180-2\alpha)=\frac{ML}{MK}\implies ML=MK\sin2\alpha=\frac {x\sin2\alpha}6$$

Finally, observe that

$$\frac12\sqrt{S\sin2\alpha}=\frac12\sqrt{\frac14x^2\sin^22\alpha}=\frac14 x\sin2\alpha$$

You can see I get a different result, so either I am wrong (check the above ) or else it should be $\;\frac13\;$ instead of $\;\frac12\;$ in your formula...

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    Actually, I used very similar logic and also didn't get the given answer. That's why I posted my question. I think everything in your solution is true and the given answer may be wrong or there is something different that we miss. – 2017-02-05
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    @SuperMan Well, let's give it some time. Where did you get this problem from? Perhaps the source has a mistake... – 2017-02-05
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Let $h$ is an altitude to $AB$.

Thus, $AC=\frac{h}{\sin\alpha}$ and $BC=\frac{h}{\cos\alpha}$, which says that $$S=\frac{h^2}{2\sin\alpha\cos\alpha}=\frac{h^2}{\sin2\alpha},$$ which gives that our distance is $\frac{1}{3}h=\frac{1}{3}\sqrt{S\sin2\alpha}$.

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    (1) How come $\;BC=\frac h{\cos\alpha}\;$ ? (2) Why would $\;\frac13h\;$ be the distance between the **centroid** (the intersection of the three *medians*) and the hypotenuse? Perhaps you were thinking of the **orthocenter** = the intersection point of the three *heights*? – 2017-02-05
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    @DonAntonio $BC=\frac{h}{\cos\alpha}$ because $\cos\alpha=\frac{h}{BC}$. The answer on your second question follows from similarity. – 2017-02-05
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    Sorry, but I can't see it: remember that *you* defined $\;h\;$ to be the **altitude to** $\;AB\;$ , so in the little right triangle formed by this altitute, we'd have that $\;\cos\alpha=\frac{AL}{AC}\;,\;\;L=$ the point on $\;AB\;$ the altitude intersects. Even if you go to *the other* little right triangle forme by that height, we'd get $\;\cos\alpha=\sin(90^\circ-\alpha)=\frac{BL}{BC}\;$ , with $\;L\;$ the same as before. But anyway: how is the distance from the **centroid** to the hypotenuse equal to $\;\frac h3\;$ ?? – 2017-02-05
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    @DonAntonio Let $CD$ is an altitude and $CO$ is a median, $M$ is a centroid and $MK$ our distance. Since $OM=\frac{1}{3}CO$, we get $MK=\frac{1}{3}CD=\frac{1}{3}h$. – 2017-02-05
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    I just can't see how $\;MK=\frac13 CD\;$ . What we know because the medians theorem is that $\;MO=\frac13\color{red}{CO}\;$ . You seem to believe that $\;CO=CD\;$ , from which it'd follow that $\;\frac13CO=\frac13CD\;$ , I don't know why. You can check in a diagram...and of course, $\;M\;$ is on $\;\color{red}{CO}\;$ , **not** on $\;CD\;$ ! – 2017-02-05
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    @DonAntonio $\Delta ODC\sim\Delta OKM$ – 2017-02-05
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    True, but then you should use a *similarity* proportion, not a direct one. Am I missing something here? – 2017-02-05
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    @DonAntonio $\frac{MK}{h}=\frac{MO}{CO}=\frac{1}{3}$ – 2017-02-05