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Interesting geometry
AlexCenteno2007   1
N 6 minutes ago by MathLuis
Let the circle (I) and G be the inscribed circle (with center I) and the centroid of triangle ABC, respectively.
X and X' are the points of tangency of the inscribed circle A-mixtiline and the eccentric circle A-mixtiline with side BC of triangle ABC, respectively.
M is the midpoint of side BC.
K is the intersection of the incircle of circle (I) with BC.
AN is the height of triangle ABC from vertex A.
K' is the reflection of K with respect to M.
L is the second point of intersection between circle ABC (circle of the triangle) and the reflection of line AM with respect to line AI.

NG, XK, LM, and X'K' meet on the circle of ABC.
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AlexCenteno2007
11 minutes ago
MathLuis
6 minutes ago
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comp. geo starting with a 90-75-15 triangle. <APB =<CPQ, <BQA =<CQP.
parmenides51   1
N Apr 6, 2025 by Mathzeus1024
Source: 2013 Cuba 2.9
Let ABC be a triangle with $\angle A = 90^o$, $\angle B = 75^o$, and $AB = 2$. Points $P$ and $Q$ of the sides $AC$ and $BC$ respectively, are such that $\angle APB = \angle CPQ$ and $\angle BQA = \angle CQP$. Calculate the lenght of $QA$.
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parmenides51
Sep 20, 2024
Mathzeus1024
Apr 6, 2025
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comp. geo starting with a 90-75-15 triangle. <APB =<CPQ, <BQA =<CQP.
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Source: 2013 Cuba 2.9
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parmenides51
30629 posts
parmenides51
#1 h  Sep 20, 2024, 9:25 PM
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Let ABC be a triangle with $\angle A = 90^o$, $\angle B = 75^o$, and $AB = 2$. Points $P$ and $Q$ of the sides $AC$ and $BC$ respectively, are such that $\angle APB = \angle CPQ$ and $\angle BQA = \angle CQP$. Calculate the lenght of $QA$.
This post has been edited 2 times. Last edited by parmenides51, Sep 20, 2024, 9:37 PM
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Mathzeus1024
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Mathzeus1024
#2 h  Apr 6, 2025, 11:17 AM
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Given right $\Delta ABC$ described above, let us examine the triangles $\Delta ABQ, \Delta APQ,$ and $\Delta CPQ$. If $\angle{APB}=\angle{CPQ}=\alpha$ and $\angle{BQA}=\angle{CQP}=\beta$, then we obtain $\alpha + \beta = 165^{\circ}$ (i) from $\Delta CPQ$. Next, an application of Law of Sines to $\Delta ABQ$ and $\Delta APQ$ yields the following system of equations in $AQ$ and $\alpha$:

$\frac{AB}{\sin(\beta)} = \frac{AQ}{\sin(75^{\circ})} \Rightarrow \frac{2}{\sin(165^{\circ}-\alpha)} = \frac{AQ}{\sin(75^{\circ})}$ (ii);

$\frac{AP}{\sin(180^{\circ}-2\beta)} = \frac{AQ}{\sin(180^{\circ}-\alpha)} \Rightarrow \frac{2\cot(\alpha)}{\sin(2\alpha-150^{\circ})} = \frac{AQ}{\sin(\alpha)}$ (iii)

of which we ultimately obtain $\textcolor{red}{AQ = \frac{\sqrt{19+9\sqrt{3}}}{3}}$.
This post has been edited 1 time. Last edited by Mathzeus1024, Apr 6, 2025, 11:54 AM
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