X-Post

        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.

  

      ITG

 kk

x-post

solution

solutie calculatorie

generalizare la dreptunghi

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 $\measuredangle{1}=60^{\circ}$