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I proved this property with an approach involving vectors. However, there should be a much simpler, elegant geometric proof, probably utilising a bunch of angles. Here is a diagram exemplifying the property:

Orthocentre-Incentre property

And here is my very tedious approach:

Let $ \lambda_1 = \frac{ \overrightarrow{OA} \cdot \overrightarrow{OB}}{\overrightarrow{OA} \cdot \overrightarrow{OA}} = \frac{|\overrightarrow{OB}|}{|\overrightarrow{OA}|} \cos(\theta) $ and $ \lambda_2 = \frac{ \overrightarrow{OA} \cdot \overrightarrow{OB}}{\overrightarrow{OB} \cdot \overrightarrow{OB}} = \frac{|\overrightarrow{OA}|}{|\overrightarrow{OB}|} \cos(\theta) $

By considering the intersection of the lines $BP_1$ and $AP_2$, where

$$ AP_2 := (1-\mu_1)\lambda_1 \overrightarrow{OA} + \mu_1 \overrightarrow{OB} \\ BP_1 := (1-\mu_2)\lambda_2 \overrightarrow{OB} + \mu_2 \overrightarrow{OA} \\ (1-\mu_1)\lambda_1 \overrightarrow{OA} + \mu_1 \overrightarrow{OB} = (1-\mu_2)\lambda_2 \overrightarrow{OB} + \mu_2 \overrightarrow{OA} \\ \implies ((1-\mu_1)\lambda_1 - \mu_2 )\overrightarrow{OA} = ((1-\mu_2)\lambda_2 - \mu_1 )\overrightarrow{OB} \\ \implies (1-\mu_1)\lambda_1 = \mu_2 \quad \text{ and } \quad (1-\mu_2)\lambda_2 = \mu_1 \\ \implies \mu_1 = \frac{\lambda_2 - \lambda_1\lambda_2}{1-\lambda_1\lambda_2} \quad \text{ and } \quad \mu_2 = \frac{\lambda_1 - \lambda_1\lambda_2}{1-\lambda_1\lambda_2} \\ \ \\ \ \\ \implies \overrightarrow{OX} = \frac{1}{1-\lambda_1\lambda_2} \left( (\lambda_1 - \lambda_1\lambda_2)\overrightarrow{OA} + (\lambda_2 -\lambda_1\lambda_2) \overrightarrow{OB} \right) $$

Similarly, by considering the intersection of $AB$ and $OX$,

$$ \overrightarrow{OP_3} = \frac{1}{\lambda_1 + \lambda_2 - 2\lambda_1\lambda_2} \left( (\lambda_1 - \lambda_1\lambda_2)\overrightarrow{OA} + (\lambda_2 -\lambda_1\lambda_2)\overrightarrow{OB} \right) \\ \overrightarrow{P_1P_3} = \frac{1-\lambda_1}{\lambda_1 + \lambda_2 - 2\lambda_1\lambda_2} \left( (\lambda_1 - 2\lambda_1\lambda_2)\overrightarrow{OA} + \lambda_2\overrightarrow{OB} \right) \\ \overrightarrow{P_2P_3} = \frac{1-\lambda_2}{\lambda_1 + \lambda_2 - 2\lambda_1\lambda_2} \left( \lambda_1\overrightarrow{OA} + (\lambda_2 - 2\lambda_1\lambda_2)\overrightarrow{OB} \right) \\ $$

Since we are looking for angles, the magnitudes of the vectors do not matter; we may strip the constants away to simplify our dot product.

$$ \overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_1P_{3\text{ dir}}} = |\overrightarrow{OP_{3\text{ dir}}}| |\overrightarrow{P_1P_{3\text{ dir}}}| \cos(\alpha_1) = \ldots = \sin^2(\theta)\cos^2(\theta) (\overrightarrow{OA} - \overrightarrow{OB})^2 \\ \overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_2P_{3\text{ dir}}} = |\overrightarrow{OP_{3\text{ dir}}}| |\overrightarrow{P_2P_{3\text{ dir}}}| \cos(\alpha_2) = \ldots = \sin^2(\theta)\cos^2(\theta) (\overrightarrow{OA} - \overrightarrow{OB})^2 \\ \implies \frac{\overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_1P_{3\text{ dir}}}}{\overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_2P_{3\text{ dir}}}} = \frac{|\overrightarrow{P_1P_{3\text{ dir}}}|\cos(\alpha_1)}{|\overrightarrow{P_2P_{3\text{ dir}}}|\cos(\alpha_2)} = 1 $$

Finally, by expansion, note that $\frac{|\overrightarrow{P_1P_{3\text{ dir}}}|}{|\overrightarrow{P_2P_{3\text{ dir}}}|} = \frac{\left| |\overrightarrow{OB}| \widehat{OA} - 2\overrightarrow{OA}\cos(\theta) + |\overrightarrow{OA}| \widehat{OB}\right|}{ \left| |\overrightarrow{OB}| \widehat{OA} - 2\overrightarrow{OB}\cos(\theta) + |\overrightarrow{OA} \widehat{OB}\right|} $, which can then be geometrically proved to be $\frac{|\overrightarrow{OA}-\overrightarrow{OB}|}{|\overrightarrow{OA}-\overrightarrow{OB}|} = 1$, here is another diagram:

Proving a ratio to be 1

Therefore, $\cos(\alpha_1) = \cos(\alpha_2) \implies \alpha_1=\alpha_2 $.

Now you see why I think there must be a much easier method. How else can we prove this?

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    $\begingroup$ It's helpful to remember —and easy to prove— that joining the feet of perpendiculars creates a triangle similar to the original. So, $\triangle OAB \sim \triangle OP_2P_1 \sim \triangle P_3AP_1\sim \triangle P_3 B P_2$. Consequently, $\cong \angle AP_3 P_1 \cong \angle O \cong \angle BP_3 P_2$, and we deduce $\alpha_1=\alpha_2$. Etc. $\endgroup$ – Blue Jun 6 '18 at 20:16
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$OP_1XP_2$ and $P_1AP_3X$ are cyclic so $$\angle P_2P_1X = \angle P_2OX= \angle P_2AB =\angle P_3P_1X $$ so $P_1B$ is angle bisector for $\angle P_3P_1P_2$...

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  • $\begingroup$ $\angle P_2P_1X=\angle P_2OX=90^{\circ}-\angle P_2XO=90^{\circ}-\angle AXP_3=\angle P_3AX=\angle P_3P_1X$. $\endgroup$ – Rosie F Jan 20 '19 at 14:48

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