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10 votes
Accepted

The maximum area of a pentagon inside a circle

Consider any two consecutive sides $AB$, $BC$ of a convex polygon inscribed in a given circle. Suppose to keep all vertices fixed apart $B$: the area of the polygon is maximum when $B$ is the midpoint ...
Intelligenti pauca's user avatar
4 votes
Accepted

Find the angle $x$ in the triangle $APQ $ below

As in the figure below, let $R$ be the third vertex of the equilateral triangle $AQR$ (with $R$ on the same side as $B$, with respect to $AC$). $ARB$ is isosceles by construction and the hypotheses, ...
dfnu's user avatar
  • 7,613
3 votes

The maximum area of a pentagon inside a circle

Let's look at general (convex) $n$-sided polygons inscribed in a unit circle. Join the centre of the circle to each vertex of the polygon to form $n$ isosceles triangles. Say their apex angles are $\...
Chris Lewis's user avatar
  • 2,618
2 votes
Accepted

Find the minimum value of the sum of the squares of the distances from M to the lines AB, AC and BC

As you said, the sum of the three squared distances, by Pythagorean theorem are $ f = 3 (1)^2 + (M'X)^2 + (M'Y)^2 + (M'Z)^2 $ Now we know that these altitudes satisfy $ \frac{1}{2} ( 2 \sqrt{3} ) ( M'...
Quadrics's user avatar
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2 votes

Find the segment "DC" in the obtuse triangle below

Let $R$ be the midpoint of $AQ$, so $\lvert AR\rvert = \lvert RQ\rvert = 5$. Since $\triangle ADQ$ is right-angled, then $AQ$ is the diameter of its circumcircle, so $R$ is its center, which means ...
John Omielan's user avatar
  • 49.7k
2 votes
Accepted

Find the segment "DC" in the obtuse triangle below

Let $\angle BAD = \angle CAD = \alpha$ then $\angle BAD =\angle BCA = 2\alpha$ Draw $DE$ such that $AE=QE$ ($E$ is the midpoint of $AQ$) then $AE=QE=DE=5$ and $\angle ADE = \angle DAE = \alpha$ then $\...
Lion Heart's user avatar
  • 7,468
1 vote

Find the segment "DC" in the obtuse triangle below

Using your result of $\measuredangle CQD = 90^{\circ} + \theta$ and the Law of sines with $\triangle CDQ$ gives $$\begin{equation}\begin{aligned} \frac{\lvert DC\rvert}{\sin(90^{\circ} + \theta)} &...
John Omielan's user avatar
  • 49.7k
1 vote

How to express an angle between two angle bisectors in interior angles of a convex quadrilateral?

$\angle LEC = \angle BEA = 180^{\circ} - \frac{\angle A}{2} - \angle B$ $\alpha + 180^{\circ} - \frac{\angle A}{2} - \angle B + \frac{180^{\circ} -\angle C}{2} = 180^{\circ}$ (in $\triangle LEC$) $\...
Lion Heart's user avatar
  • 7,468

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