ogeometrie: februarie 2022

miercuri, 23 februarie 2022

GAZETA MATEMATICĂ Seria B N0 1/2022

vezi in DRIVE

Vezi ERATA

SUPLIMENTUL cu EXERCIȚII al GMB N0 1/2022

vezi in DRIVE

joi, 17 februarie 2022

Karede Bir Açı ve Bir ... Hiçbir Yerden Uçurtma

An Angle in a Square and a Kite Out of Nowhere

Uçurtma, akademik dilde Deltoid olarak da adlandırılır.

Starting from a geometry problem and the solution that Hasan Ata offered, we formulate the following problem:

" Let $ABCD$ be a square and an angle $\angle EAF$ with $E \in ]BC[,\;F \in ]CD[$ . Prove that

$\measuredangle EAF=45^{\circ}\;\Leftrightarrow\;EF=BE+DF."$

Solution CiP

$\measuredangle EAF=45^{\circ}$

Rotate the $ABE$ with $90^{\circ}$ clockwise around point $A$ and obtain the triangle $ADG$ . We have $\measuredangle EAG=90^{\circ}$ , $AE=AG$ and $DG=BE$ .

Since $\measuredangle FAG=90^{\circ}-45^{\circ}=45^{\circ}=\measuredangle EAF$ , it follows that the line $AF$ is the angle bisector of the $\angle EAG$ in the isosceles triangle $AEG$ so it is the perpendicular bisector of the segment $[EG]$ , hence $FG=FE$ , so $EF=FG=DG+DE=BE+DF$ .

q.e.d.

$EF=BE+DF$

The same rotation now shows us that $FG=FE$ so quadrilateral $AEFG$ is a kite.

Hence line $FA$ is simultaneously angle bisector of $\angle EFG$ and $\angle EAG$ . Because $\measuredangle EAG=90^{\circ}$ it follow $\measuredangle EAF=\frac{90^{\circ}}{2}=45^{\circ}.$

q.e.d.

$\blacksquare$

miercuri, 16 februarie 2022

DIDACTICA MATEMATICĂ N0 1/2021

vezi in DRIVE

marți, 15 februarie 2022

Ten Solutions to a Geometry Problem

This is the article signed by N. Mihăileanu in

at pages 202-208.

DIDACTICA MATEMATICĂ N0 2/2020

vezi in DRIVE

luni, 14 februarie 2022

Три проблема с приблизително една и съща конфигурация

Това са броеве 1334, 1363 и 1366, публикувани тук http://gogeometry.com/.

Problem 1334 (reformulation)

Let $E$ be a point on the side $]BC[$ of the square $ABCD$ . $[AF],\;\;F \in ]CD[$ is the bisector of the $\angle DAE$ . Prove that $AE=BE+DF$ .

In relation to the original figure, with the changed notes

we have equal angles

$\angle BCE = \angle DHE$ (from the parallelism of the

$CB$ and

$AH$ lines), and

$\angle BCE=\angle EBG$ , hence

$\angle EBG = \angle GHB$ so

$BGH$ is an isosceles triangle and then

$BG=a$ .

Let's get back to our figure.

$ADG$ so that $\Delta ABE \equiv \Delta ADG$ (basically, we rotate the triangle $ABE$ with $90^{\circ}$ around point $A$ , counterclockwise), we have $\measuredangle AGD=x^{\circ}$ and $[AG] \equiv [AE]$ .

$AGF$ : $\measuredangle AFG=180^{\circ}-\measuredangle AGF-\measuredangle GAF=180^{\circ}-x^{\circ}-(90^{\circ}-\frac{x^{\circ}}{2})=90^{\circ}-\frac{x^{\circ}}{2}$ .

From $\angle AFG \equiv \angle GAF$ triangle $AFG$ is isosceles, with $GA=GF=a+b$ .

$\blacksquare$

Problem 1363

Solution CiP

From $FC \perp BC,\;FG\perp BE$ and fact that $BF$ is bisector of $\angle CBE$ it follow $FC=FG$ so the congruence $\Delta BCF \equiv \Delta BGF$ and hence $BG=BC$ . But $AB=BC$ so triangle $ABG$ is isosceles.

Let $x^{\circ}=\measuredangle CBF=\measuredangle FBG$ , then $\underline{\measuredangle GFB}=\measuredangle CFB=\underline{90^{\circ}-x^{\circ}}$ .

Now, $\measuredangle ABG=90^{\circ}-2x^{\circ}$ , so $\measuredangle BAG=\underline{\measuredangle BGA}=\underline{45^{\circ}+x^{\circ}}$ .

We calculate now $\measuredangle FGH=180^{\circ}-\measuredangle FGA=180^{\circ}-\measuredangle FGB-\measuredangle BGA=$

$=180^{\circ}-90^{\circ}-(45^{\circ}+x^{\circ})=45^{\circ}+x^{\circ}$ .

But $\measuredangle BFG=\measuredangle FHG +\measuredangle FGH$ (as the outer angle of the triangle FGH) so

$\measuredangle FHG=\measuredangle BFG-\measuredangle FGH=(90^{\circ}-x^{\circ})-(45^{\circ}-x^{\circ})=45^{\circ}$ .

$\blacksquare$

Problem 1366

Solution CiP

Consider the square $ABCD$ and the point $E$ on $]AD[$ .

Line

$BE$ intersects the circle once again at point

$E'$ and the bisector

$BF$ of the angle

$\angle CBE$ intersects the circle again at point

$F'$ . Let

$BE \cap AF' ={J}$ .

We will prove that $[BA] \equiv [BJ]$ , so $J=G$ and so $F'=H$ . (The fact that $BG$ is equal to the side of the square results easily from the congruence of the triangles $\Delta BCF \equiv \Delta BGF$ .)

First notice that point $F'$ is the middle of the arc $\overset{\frown}{C_DE'}\;:\;\overset{\frown}{CF'} \equiv \overset{\frown}{F'E'}$ .

Now applying the calculation formulas of the angles inscribed in the circle, respectively the angle with the tip inside the circle, we obtain

$\measuredangle BAJ=\measuredangle BAF'=\frac{\overset{\frown}{B_CF'}}{2}=\frac{\overset{\frown}{BC}+\overset{\frown}{CF'}}{2}=\frac{90^{\circ}+\overset{\frown}{F'E'}}{2}=\frac{\overset{\frown}{AB}+\overset{\frown}{E'F'}}{2}=\measuredangle BJA.$

It turns out that the triangle $ABJ$ is isosceles, with $BA=BJ$ .

$\blacksquare$

Remark : $\measuredangle AHB=\frac{\overset{\frown}{AB}}{2}=\frac{90^{\circ}}{2}=45^{\circ}.$