Hi ; I Don't Understand What Is Mean Distinct Right Angles , You Mean For The $n=2$ There Is $6$ Distinct Right Angles Or $7$ ??? Best Regard

War-Hammer wrote: Hi ; I Don't Understand What Is Mean Distinct Right Angles , You Mean For The $n=2$ There Is $6$ Distinct Right Angles Or $7$ ??? Best Regard it depends on formation of rectangles. for example, consider two rectangles with following coordinates: $(0,0),(0,1),(1,0),(1,1)$ and $(0,0),(0,2),(2,0),(2,2)$ in this example we have $7$ distinct right angles. note: different rectangles sides are not necessarily parallel. for example: $(0,1),(0,-1),(1,0),(-1,0)$ is possible.

Hi ; Solution : ( This Is My Friend Solution ) (Solution For Square) First It's Obvious There Is No Two Angle Which Are In Two Squares , Cause Then Tow Squares Are Coinciding . Let $A$ Be The Right Angle Which Is Come To $k$ Squares Where $k$ Is Maximum . So For These $k$ Squares We Have $3k+1$ Right Angle For $n-k$ Other Squares Hence $k$ Is Maximum Then We Have Minimum $\dfrac{3(n-k)}{k}$ New Right Angle . So We Have At Least $3k+1+\dfrac{3n}{k}\ -3$ Right Angle , But We Have $3k + \dfrac{3n}{k}\geq 6\sqrt{n}$ ( AM-GM ) So We Have At Least $ 6\sqrt{n} - 2 \geq 4\sqrt{n}$ Q.E.D Best Regard

To be honest I stopped reading here: War-Hammer wrote: First It's Obvious There Is No Two Angle Which Are In Two Rectangle , Cause Then Tow Rectangle Are Coinciding How about the rectangles with corners $\{(0,0), (0,1), (1,1),(1,0)\}$ and $\{(0,0),(0,1),(2,1),(2,0)\}$? (Also, about conventions of written English: typically only the first word in each sentence has its first letter capitalized; there are a special class of nouns (the proper nouns) that are capitalized whenever they appear, but in your post there are at most one or two of these.)

JBL wrote: To be honest I stopped reading here: War-Hammer wrote: First It's Obvious There Is No Two Angle Which Are In Two Rectangle , Cause Then Tow Rectangle Are Coinciding Hi ; The Second One Is Rectangle , I Wrote , I Solve The Problem For Squares Best Regard

We can show that there are at least $4\sqrt{n}$ distinct right angles. Induct on $n$, $n=1$ case is obvious. If there are two rectangles that are not parallel, we can finish the problem by using induction hypotehsis and $\sqrt{a}+\sqrt{b} \geq \sqrt{a+b}$. Suppose all rectangles are parallel. Extending the sides of each rectangle, we can make a grid. Assume there are $k$ horizontal lines. In the $i$'th horizontal line(from top to bottom), denote the number of distinct right angles touching the line and heading bottom-left by $a_i$, and right angles touching the line and heading top-right by $b_i$. Each rectangle has a right angle heading bottom-left and top-right, so $\sum_{i<j}a_i b_j \geq n$. $n \leq \sum_{i<j}a_i b_j \leq (\sum a_i )(\sum b_i ) \leq \frac{(\sum a_i + \sum b_i )^2}{4}$, therefore $\sum a_i + \sum b_i \geq 2\sqrt{n}$. Working similarly with right angles heading bottom-right and top left, we get that the total number of right angles are at least $4 \sqrt{n}$.