Não importa o quanto as Áreas se escondam, ainda as encontraremos // No matter how much the Areas hide, we will still find them

           Based on a LinkedIn Post

Then I found a similar one

I thought of a general formula, resulting in the following:

          PROBLEM In the parallelogram $ABCD$ in the figure below, we denote the

                        areas of the triangles $ABM,\;KLM$ by $\textbf{S}_1,\; \textbf{S}_2$ and the areas of the

                        quadrilaterals $ADLM,\;BCKM$ by $\textbf{S}_3,\;\textbf{S}_4$. The following                                              relationship holds:

$$\textbf{S}_3+\textbf{S}_4=\textbf{S}_1-\textbf{S}_2+2 \cdot \sqrt{\textbf{S}_1 \cdot \textbf{S}_2} \tag{1}$$

          Solution CiP

               I added additional lines and notations to the figure, compared to those mentioned in the statement: $$EF \parallel AB,DC,\;EF \ni M,\;\;PQ\perp AB,DC,\;PQ \ni M.$$

          It is known that the ratio of the areas of two similar triangles

$\frac{\textbf{S}_2}{\textbf{S}_1}=\left ( \frac{MQ}{MP} \right )^2\;\;\Rightarrow\;\frac{MQ}{MP}=\sqrt{\frac{\textbf{S}_2}{\textbf{S}_1}} \tag{2}$

We also have $\frac{\textbf{S}_{CEFD}}{\textbf{S}_{ABEF}}=\frac{EF \cdot MQ}{EF \cdot MP}=\frac{MQ}{MP}\underset{(2)}{=}\sqrt{\frac{\textbf{S}_2}{\textbf{S}_1}}\;\;\;\Rightarrow \;\textbf{S}_{CEFD}=\textbf{S}_{ABEF} \cdot \sqrt{\frac{\textbf{S}_2}{\textbf{S}_1}} \tag{3}$

     On the other hand $\textbf{S}_{ABEF}=AB \cdot MP=2\cdot \textbf{S}_{ABM}=2\cdot\textbf{S}_1\;\;\overset{(3)}{\Rightarrow}\;\textbf{S}_{CEFD}=2\cdot \sqrt{\textbf{S}_1\cdot \textbf{S}_2}$

and      $\textbf{S}_{ABCD}=\textbf{S}_{ABEF}+\textbf{S}_{CEFD}=2\cdot \textbf{S}_1+2\cdot \sqrt{\textbf{S}_1 \cdot \textbf{S}_2}\;\;\;\Leftrightarrow$

$\Leftrightarrow \textbf{S}_1+\textbf{S}_2+\textbf{S}_3+\textbf{S}_4=2\cdot \textbf{S}_1+2\cdot \sqrt{\textbf{S}_1 \cdot \textbf{S}_2}\;\;\;\Rightarrow \textbf{S}_3+\textbf{S}_4=\textbf{S}_1-\textbf{S}_2+2\cdot \sqrt{\textbf{S}_1 \cdot \textbf{S}_2}$

QED $\blacksquare$

Tiger oder fatiguer Geometry ???

        Find the area of ​​the parallelogram $ABCD$

ANSWER CiP            $\textbf{S}_{ABCD}=96$

    

          The problem is from Igor's Twitter post. He also gave a solution. In a later comment he said that the original Square could only be a Rectangle, refraining from the same answer. The author of Peter Gallin gave a much appreciated computational solution, which probably remains valid for rectangles as well. I analyzed Igor's solution, which is not explained in detail enough. He writes some equations that lead to the result. Even for this he deserves our thanks.    

   

                                           Solution CiP

              I appreciate that Igor's solution is based on the geometric results which also appear in my Post A LEMMA and a COROLLARY ... It will be seen that in this way the Problem can be formulated even in a Parallelogram. The calculations made by Peter Gallin helped me draw a corresponding figure to scale. We thank him too.

               We introduce additional notations in the figure:
points $G,\;H,\;K$ and segments $[BG],\;[DG].$ We also call the areas $\textbf{S}_{AFG}=U,\;\textbf{S}_{CEK}=W.$

          Applying the Lemma from the previously mentioned post to the quadrilaterals $ADGF,\;BCEG$ we have that the following areas have the values 

$$\textbf{S}_{DGH}=\frac{180}{U},\quad \quad \textbf{S}_{BGK}=\frac{9}{W} \;.\tag{1}$$

          It remains to express the areas marked with $?,\;??$ in the figure. For this we will use another result, valid in parallelograms, namely that, in the adjacent figure

 we have $\textbf{S}_{ABG}=\textbf{S}_{ADC}$ (and of course $\textbf{S}_{BCG}=\textbf{S}_{CDG}$). Apologies for overlapping some notations !! Indeed, for $BH,\;DJ \perp AC$ we have, from congruent triangles $\Delta ADJ \equiv \Delta CBH$ that $BH=DJ$ and sos

$$2 \cdot \textbf{S}_{ABG}=AG \cdot BH=AG \cdot DJ=2 \cdot \textbf{S}_{ADG}$$

 

          Returning to the figure from the problem, we will apply this situation exactly to point $G$, and so we can write 

$$\textbf{S}_{ABG}=\textbf{S}_{ADG}\quad \quad \textbf{S}_{CDG}=\textbf{S}_{BCG}\;\;\; \Leftrightarrow$$

$$\Leftrightarrow \;\;\; U+10+?=18+\frac{180}{U} \quad \quad W+1+??=9+\frac{9}{W}\;\;\; \Leftrightarrow$$

$$\Leftrightarrow \;\;\;\;?=\frac{180}{U}-U+8 \quad \quad ??=\frac{9}{W}-W+8 \tag{2}$$

Thus, I found all the information presented in the figure from Igor's solution. (I changed the letters A and B to W and U). All that remains is to write the equations that solve the problem.

          We will apply the Corollary we mentioned at the beginning for trapezoids $ABCE$ and $ADCF$ (Note that we don't even have to draw them in full because we would unnecessarily clutter up the figure!) Namely

 $\textbf{S}_{BCK}^2=\textbf{S}_{CEK} \cdot \textbf{S}_{ABK}\quad \quad \textbf{S}_{ADH}^2=\textbf{S}_{AFH} \cdot \textbf{S}_{DCH} \tag{3}$

But from the figure with the areas marked on it we also find

$\textbf{S}_{ABK}=U+10+?+\frac{9}{W}\overset{(2)}{=}U+10+\frac{180}{U}-U+8+\frac{9}{W}=\frac{9}{W}+\frac{180}{U}+18$

$\textbf{S}_{CDH}=W+1+??+\frac{180}{U}\underset{(2)}{=}W+1+\frac{9}{W}-W+8+\frac{180}{U}=\frac{9}{W}+\frac{180}{U}+9$

and then the relations (3) are written

$$9^2=W \cdot \left ( \frac{9}{W}+\frac{180}{U}+18 \right )\quad \quad 18^2=U \cdot \left ( \frac{9}{W}+\frac{180}{U}+9\right ) \tag{4}$$

From here we will find $U$ and $W$. Let us also note that the area of ​​the entire parallelogram is

$2\cdot \textbf{S}_{ABC}=U+10+?+\frac{9}{W}+9=\underset{(2)}{=}U+10+\frac{180}{U}-U+8+\frac{9}{W}+9$ so
$$\textbf{S}_{ABCD}=2 \cdot \left (\frac{9}{W}+\frac{180}{U}+27 \right ) \tag{5}$$

          We will leave the geometry aside and move on to solving the system of equations (4). After performing the calculations and simplifications, it is written

$$\begin{cases}10 \frac{W}{U}+W=4\\\frac{U}{W}+U=16\end{cases}$$

Eliminating the denominators will result

$$\begin{cases}UW=4U-10W\\UW=16W-U \end{cases}$$

From here $4U-10W=16W-U\;\Leftrightarrow\;5U=26W$ and substituting $\frac{U}{W}=\frac{26}{5}$ into one of the equations we have $\frac{26}{5}+U=16\;\;\Rightarrow U=\frac{54}{5}$. Then $W=\frac{5}{26}\cdot U=$ $=\frac{5}{26}\cdot \frac{54}{5}=\frac{27}{13}.$

          Oh, finally, from (5) we calculate the area of ​​the parallelogram
$$\textbf{S}_{ABCD}=2 \cdot \left (\frac{9}{\frac{27}{13}}+\frac{180}{\frac{54}{5}}+27  \right )=96$$

$\blacksquare$

ЛЕМА и ЗАКЛУЧОК кои не се премногу НЕКОРИСНИ // A LEMMA and a COROLLARY that are not too USELESS

          We denote in these lines with $\textbf{S}_{ABC}$ the area of ​​triangle $ABC$.

          LEMMA  Let $ABCD$ be a convex quadrilateral and $O$ be the point of

                           intersection of the diagonals $AC,\;BD.$ Then

$$\textbf{S}_{AOB} \cdot \textbf{S}_{COD}= \textbf{S}_{AOD} \cdot \textbf{S}_{BOC} \tag{1}$$

                          holds.

         Proof of Lemma {it would work immediately by applying the formula for the area of ​​a triangle $\textbf{S}_{AOB}=OA \cdot OB \cdot \sin \widehat{AOB}/2$ etc. and the equality of the sines of the supplementary angles. I prefer a demonstration that would make even a parrot envious.}

          Let ${\color{Brown} {AE}},\;{\color{Brown}{CF}} \perp BD$. 

We have the formulas

$$2\textbf{S}_{AOB}={\color{Yellow}{OB}} \cdot {\color{Brown}{AE}},\;\;\;\;2\textbf{S}_{BOC}={\color{Blue}{OB}} \cdot {\color{Brown}{CF}} \tag{2}$$

$$2\textbf{S}_{COD}={\color{Green}{OD}} \cdot {\color{Brown}{CF}},\;\;\;\;2\textbf{S}_{AOD}={\color{Red}{OD}} \cdot {\color{Brown}{AE}} \tag{3}$$

Then

$4\textbf{S}_{AOB}\cdot \textbf{S}_{COD}={\color{Yellow}{OB}} \cdot {\color{Brown}{AE}} \cdot {\color{Green}{OD}} \cdot {\color{Brown}{CF}}={\color{Blue}{OB}} \cdot {\color{Brown}{CF}} \cdot {\color{Red}{OD}} \cdot {\color{Brown}{AE}}=4\textbf{S}_{BOC} \cdot \textbf{S}_{AOD}.$

The Lemma is proven.

$\square$(QED Lemma)

          COROLLARY  In the trapezoid $ABCD,\;\; AB \parallel CD$ we have

$$\textbf{S}_{AOD}^2=\textbf{S}_{BOC}^2=\textbf{S}_{AOB} \cdot \textbf{S}_{COD} \tag{4}$$

          Indeed, let $AE,\;BF \perp CD$. We have $AE=BF$, so

$2\textbf{S}_{ACD}=AE \cdot CD=BF \cdot CD=2 \textbf{S}_{BCD}\;\;\Rightarrow\;\textbf{S}_{ACD}-\textbf{S}_{COD}=\textbf{S}_{BCD}-\textbf{S}_{COD}\Rightarrow$

$$\Rightarrow\;\;\textbf{S}_{AOD}=\textbf{S}_{BOC}$$

and (4) results from (1).

$\blacksquare$

An Equation That Saved the World // Një ekuacion që shpëtoi botën

                Well, not exactly the World, but at least the Problem from yesterday's Post. I promised an analytical solution there, which here it is.

          We choose a Cartesian Frame with the origin at point $O$, the $x$-axis along the side $OA$ and the $y$-axis along the side $OB$. The coordinates are marked on the figure.

          We will calculate the slope of line $AF$, showing that it is $-1$ so $\phi=135^{\circ}$.

          The equation of the line $OE$, which passes through points $O(0,0)$ and $E(a-c,b)$, is

$$y=\frac{b}{a-c} \cdot x.\tag{OE}$$

The equation of the line $CD$, which passes through points $C(0,c)$ and $D(a,b)$ is

$$y-c=\frac{b-c}{a-0} \cdot (x-0). \tag{CD}$$

Point $F$ - the intersection of lines $OE$ and $CD$ - has the coordinates obtained from solving the two equations $(OE),\;(CD):$

$$\frac{b}{a-c} \cdot x-c=\frac{b-c}{a} \cdot x\;\;\Rightarrow\;x=x_F=\frac{a(a-c)}{a+b-c};\tag{$x_F$}$$

and $y_F=y\underset{(OE)}{=}\frac{b}{a-c} \cdot x_F=\frac{b}{a-c}\cdot \frac{a(a-c)}{a+b-c}=\frac{ab}{a+b-c}. \tag{$y_F$}$

      The slope of line $AF$ is

$m=\tan \phi =\frac{y_F-y_A}{x_F-x_A} \overset{(x_F)\;(y_F)}{===} \frac{\frac{ab}{a+b-c}-0}{\frac{a(a-c)}{a+b-c}-a}=\frac{\frac{ab}{a+b-c}}{\frac{a^2-ac-a^2-ab+ac}{a+b-c}}=\frac{ab}{-ab}=-1.$

$Q.E.D.$

Walking on the edges of a rectangle

      From X-post.

       

          We will post a solution with Analytical Geometry later.

ANSWER CiP

$$\alpha=45^{\circ}$$

                    

                           Solution CiP

          Let's make some notations on the figure: $OA=a,\;OB=b,\;OC=DE=c.$

The extension of the segment $OE$ beyond point $E$ intersects the line $AD$ at point $G$.

          From the similar triangles $\Delta OBE \sim \Delta GDE$ (because $GD \parallel BO$)

 we have $\frac{OB}{GD}=\frac{EB}{ED}\;\Leftrightarrow \frac{b}{GD}=\frac{a-c}{c}$, hence

$$GD=\frac{b\cdot c}{a-c} \tag{1}$$

For the same reason, we have the similar triangles $\Delta CFO \sim \Delta DFG$, so

$$\frac{FO}{FG}=\frac{CO}{GD}\underset{(1)}{=}\frac{c}{\frac{bc}{a-c}}=\frac{a-c}{b}. \tag{2}$$

          On the other hand $\frac{AO}{AG}\overset{(1)}{=}\frac{a}{b+\frac{bc}{a-c}}=\frac{a}{\frac{ab}{a-c}}=\frac{a-c}{b}.$ Comparing with (2) it is seen that

$$\frac{FO}{FG}=\frac{OA}{AG}$$

so by the conversely of the Angle Bisector Theorem it follows that $AF$ is the angle bisector of angle $\measuredangle OAG=90^{\circ}$.

$\blacksquare$

BĂDESCU Lucian - a geometer not like many others

                From X-post

And the aforementioned book by Badescu

Circle Inscribed in a Half of a SQUARE

 From a Linkedin Post

ANSWER CiP

$$\frac{a}{b}=1+\sqrt{2}$$

                    Solution CiP

          Notations

We have the equality of triangles (ASA)

$$\Delta BEO \equiv \Delta DFO$$

because $OB \equiv OD:=a$, $\measuredangle {OBE}=\measuredangle{ODF}=45^{\circ}, \; \measuredangle {BOE} \equiv \measuredangle{DOF}$ as vertical angles.

From here $DF=b$, so $CF=a$. But $CF \equiv CO$ as tangents from point C to the circle. Then, from the right and isosceles triangle $BOC$ we find $BC=a\cdot \sqrt{2}$.

          We have the equations $a+b=AB=BC=a\cdot \sqrt{2}$, hence

 $a(\sqrt{2}-1)=b\;\Rightarrow a(\sqrt{2}-1)(\sqrt{2}+1)=b(\sqrt{2}+1) \Leftrightarrow a(2-1)=b(\sqrt{2}+1)$, so $\frac{a}{b}=\sqrt{2}+1$. We got the answer.

$\blacksquare$

Примена једначине 3-степена у геометрији // Applying a 3rd degree Equation in Geometry

           From a LinkedIn post of Reygan Diionisio , see here. You can also watch it on YouTube.

ANSWER CiP

$$x=3$$

                    Solution CiP

          With the notations in the figure below

     We will apply the Law of cosine to triangles. More precisely, a consequence of it.

$$\cos \measuredangle ACB=\frac{a^2+b^2-c^2}{2ab} \tag{1}$$

where, for triangle $ABC$ we note $a=BC,\;b=AC,\;c=AB.$

Applying (1) to triangle $ABE$ we have

$$\cos \measuredangle{1}=\cos \widehat{AEB}=\frac{AE^2+BE^2-AB^2}{2\cdot AE \cdot BE}=$$

$=\frac{(x+5)^2+x^2-49}{2x(x+5)}=\frac{2x^2+10x-24}{2x(x+5)}=\frac{x^2+5x-12}{x(x+5)}. \tag{2}$

Applying (1) to triangle $CDE$ we have

$$\cos \measuredangle{2}=\cos \widehat{CED}=\frac{CE^2+ED^2-CD^2}{2\cdot CE \cdot ED}=$$

$=\frac{25+x^2-49}{2\cdot 5 \cdot x}=\frac{x^2-24}{10x}. \tag{3}$

Given that angles $\measuredangle{1}$ and $\measuredangle{2}$ are supplementary, we have
$$\cos \measuredangle{1}=-\cos \measuredangle{2}$$

so, from (2) and (3), we have the equation

$$\frac{x^2+5x-12}{x(x+5)}=-\frac{x^2-24}{10x}$$

$\Leftrightarrow\;10\cdot(x^2+5x-12)=-(x+5)(x^2-24)\Leftrightarrow x^3+15x^2+26x-240=0 \tag{4}$

We easily find that the equation (4) has the rational root $x_1=3$ through the Horner's Method:

$$\begin{array}{c|c} \;&1&15&26&-240\\ \hline 3&1&18&80&0 \end{array}$$

Also from the table, second line, we see that the other two roots are given by the equation

$x^2+18x+80=0$, but they are $x_2=-10,\;x_3=-8$ but they have no geometric significance, being negative.

$\blacksquare$

          Remark CiP  For $x=3$, from the relation (2) we obtain $\cos \measuredangle{1}=\frac{1}{2}$ so $\measuredangle{1}=60^{\circ}$