3
votes
Perpendicular segments from special points in a rectangle with ratio $BC = 2AB$
Recall the setting of the problem, since we want only $KD\perp KM$ we introduce only the points that really matter:
Problem: Let $ABML$ and $LMCD$ be two squares, which are reflected versions of each ...
- 38.7k
3
votes
Accepted
Maximum number of non-crossing edges in a 7-node network, such that nodes can be three-colored without same-colored nodes being connected
For a 7 node 3 color graph there can be a maximum of 14 edges. The general formula for the two constructions I have found for the maximum number of edges in a 3 color planar graph is:
$$E(n)=\begin{...
- 131
3
votes
Accepted
Prove that for all real $a,b,c$ holds $a^2 + b^2 + c^2 + 2 + (abc)^2 \ge 2(ab + bc + ca).$
Since $x \leqslant |x|,$ we will prove the stronger inequality
$$|a|^2 + |b|^2 + |c|^2 + 2 + |a|^2|b|^2|c|^2 \geqslant 2(|a||b| + |b||c| + |c||a|).$$
Replace $(|a|,|b|,|c|) \to (x,y,z)$ then $x,y,z \...
- 6,299
1
vote
Perpendicular segments from special points in a rectangle with ratio $BC = 2AB$
Easy to see that $N$ is orthocenter of $\triangle LFC$.
So $ \triangle FML \sim \triangle FKC$
$$\Rightarrow \frac{FM}{FK}=\frac{ML}{KC}=\frac{CD}{KC}$$
And $\angle KFM=90^\circ - \angle KCF=\angle ...
- 11
1
vote
Isosceles trapezoid and special triangles in a square with perpendicular construction
Proof of (1):
Triangles $\triangle DCE$ and $\triangle BCH$ are congruent because:
$\angle DCE = \angle BCH = 90°$ (since $BC \perp DH$) ... (1)
$DC = BC$ (sides of the square) ... (2)
$\angle EDC = \...
1
vote
Collinearity of three points in a square with perpendicular and parallel constructions
Here is my synthetic proof:
Solution
Step 1: Congruent Triangles ADE and DCH
Consider triangles $\triangle ADE$ and $\triangle DCH$.
These triangles are congruent because:
(a) Right angles from the ...
1
vote
Parallelogram formed by altitude, perpendicular construction, and angle bisector in a triangle
Here is my synthetic proof:
Solution
Step 1: In the isosceles right triangle $ADE$ (where $AE = AD$ and $\angle EAD = 90°$), we can calculate that
$$\angle AED = \frac{\angle C}{2}$$
Step 2: From the ...
1
vote
Solving $1 + x\cos(ωx) + \sin(ωx)=0$ for $x$. Is there a closed form?
Partial Answer
To begin with, let us denote by $\alpha(x)$ the angle satisfying the relation:
\begin{align*}
\cos(\alpha(x)) = \frac{x}{\sqrt{x^{2} + 1}}\quad\text{and}\quad \sin(\alpha(x)) = \frac{1}{...
- 19.5k
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