Dan BRÂNZEI : Inside is Geometry! Those who love it and those who don't are invited in!

 

Evaluarea nationala

          At the National Assessment for VIII-th grade graduates, the students received the following subjects: see the section "Subiecte si bareme" (Subjects and scales).

          Regarding the Mathematics test, the subject is in the

 "ENVIII_Matematica_2024_var_07_LRO.pdf" 

file, and the correction scale in the 

"ENVIII_Matematica_2024_bar_07_LRO.pdf"

file. There was also, like every year, a reserve option : see "Matematică rezervă - 27 iunie 2024" section; see files "ENVIII_Matematica_2024_var_02_LRO.pdf" and "ENVIII_Matematica_2024_bar_02_LRO.pdf".

          Two problems were considered more difficult. Problem III.4 and problem III.5.

In translation: 

                         "The attached figure shows the isosceles triangle $ABC$ with $AB=AC$.

                       The height from the top $A$ intersects the side $BC$ at point $D$ and $AD=BC$. 

                       The height from vertex $B$ intersects side $AC$ at point $E$. Heights $AD$

                        and $BE$ intersect at point $H$.

                          a) Show that angles $DAC$ and $EBC$ have the same measure.

                          b) Prove that $AH=3 \cdot HD$."

In translation:

                         "The attached figure shows the circle with center $O$, in which $CD$ is

                          the diameter. Point $B$ belongs to the circle so that lines $BO$ and $CD$

                          are perpendicular. Point $M$ belongs to the small arc $BC$, lines $DM$

                          and $BO$ intersect at point $N$, $DN=2 \cdot MN$ and $MN=4\;cm$.

                               a)  Show that the measure of the angle $CMD$ is $90^{\circ}$.

                               b)  Calculate the area of ​​the triangle $DON$."

               The full Subject and Scale are in the Images below.

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                    Solution and comments CiP

                The solution to exercise III.4.a) is standard, noting that the angle $\measuredangle DAC$ is, in the right triangle $ACD$, the complement of the angle $\measuredangle C$, and the angle $\measuredangle EBC$ is, in the right triangle $BCE$, the complement of the same angle $\measuredangle C$.

          In exercise III.4.b) the key to the solution is the consideration of similar triangles $ACD$ and $BDH$. The order of the vertices in the relation of similarity is very important: respectively equal angles are opposed to respectively proportional sides. In more detail,

          first, $BD=DC=\frac{BC}{2}=\frac{AD}{2}$;

          secondly $\measuredangle BDH =90^{\circ}=\measuredangle ADC$ and 

$$\measuredangle HBD \underset{same\;angle}{=}\measuredangle EBC \underset{a)}{=}\measuredangle CAD.$$

Then from the last row above we have based on the "AA" criterion that

$$\Delta \;BDH \sim \Delta \;ADC$$

hence

$$\frac{BD}{AD}=\frac{DH}{DC}=\frac{BH}{AC}. \tag{$\sigma$}$$

But $\frac{BD}{AD}=\frac{\frac{BC}{2}}{BC}=\frac{1}{2}$, so from ($\sigma$) $\frac{HD}{DC}=\frac{1}{2}$. From here $HD=\frac{DC}{2}=\frac{BC}{4}=\frac{AD}{4}$.

          We get $AH=AD-HD=AD-\frac{AD}{4}=3 \cdot \frac{AD}{4}=3 \cdot HD.$

$\square$

               The key to exercise III.5.b) consists in determining the radius of the circle. For this, all the solutions I have come across are based on writing segments ratios. The most common is $\frac{OD}{ND}=\frac{MD}{CD}$. In the official scale, it results from writing the cosine of angle $\measuredangle C$ in two different triangles. Or one could use, less common with our students, the similarity of the triangles $DNO$ and $DCM$.

          But a simple idea can solve the problem without resorting to the most elementary knowledge of geometry. Join points C and M.

          $BO$ is the median of the segment $[CD]$, so the point $N$ on it verifies $NC=ND$. Hence 

$$NC=8.$$

          The right triangle $CMN$ has the leg $MN$ half of the hypotenuse $CN$ therefore it is a "30-60-90" triangle. (I have marked this on the figure with blue lines.) We still have $\measuredangle CND=180^{\circ}-60^{\circ}=120^{\circ}$. But in the isosceles triangle $CDN$, the median $NO$ from the vertex $N$ is also a bisector, so $\measuredangle CNO =\measuredangle DON =120^{\circ}\;:\;2=60^{\circ}$ (marked with red). It follows easily that the triangles $CNO$ and $DNO$ are of the same type. So $NO=4$ and we now have enough data to finish the problem.

$\blacksquare$

Juliusz Brzeziński Galois teori genom övningar // Galois Theory Through Exercises by Juliusz Brzeziński

                A source where you can download this book is here.

 

                I read an earlier version ("2015"), written for students. 

                 An interesting problem caught my attention: on page 102, problem 16.79 (in the "2015" version it is page 87, problem 16.40). But without indications of a solution. I have made posts with this type of problem here, here and here.

  

48-63-69 TRIANGLE : Să trăiești NAȘULE !! // Long live Godmother !!

           I used to follow this geometry blog. 

           I actually contributed to the publication of an problem (#963).

           The title is and is not a joke: I named my blog "ogeometrie" in the hope that those who will access "gogeometry" will also find me.

The figure above is a triangle with angles $\measuredangle {A}=48^{\circ},\;\measuredangle{B}=63^{\circ},\;\measuredangle{C}=69^{\circ}.$

          This triangle is the subject of Problem #1561. Today (2024-06-24) two solutions are already published. GregMarch 3, 2024 at 3:27 AM  and  Sumith PeirisMarch 5, 2024 at 11:48 AM. I also tried the solution, but no matter how much "angle chasing" I did, it was not enough. It is proven from the already published solutions that we must also appeal to a circumscribed circle.

Lemma ni Euler // Euler's lemma

 (filipineza) 

                    The name "EULER's Lemma" is given by me. This property of the circumcenter and the orthocenter is used in the demonstration of the "Circle of the Nine Points Theorem".

                    THEOREM (EULER's Lemma)

                    "We denote by O, G and H respectively the center of the 

                circumscribed circle, the center of gravity and the orthocenter

                 of a triangle ABC.

                     Let A' be the midpoint of side [BC]. We have equality

$$AH=2\cdot OA'. " \tag {e}$$

               Rem CiP

                In reality we have more, the relationship between vectors takes place

$$\overrightarrow{AH}=2 \cdot \overrightarrow{OA'}. \tag {E}$$

                    CiP Proof (following the text from the images at the beginning) 

               First we demonstrate

               Lemma 1 Two triangles with respectively parallel sides are similar.

                   Proof of Lemma_1

                  Let $A_1B_1C_1$ and $A_2B_2C_2$ be two triangles having

$$A_1B_1 \parallel A_2B_2\;,\;\;A_1C_1 \parallel A_2C_2\;,\;\;B_1C_1 \parallel B_2C_2\;. \tag{$\pi$}$$

Since [the angles] $\angle B_1A_1C_1,\;\angle B_2A_2C_2$ have respectively parallel sides, we have

$$\angle A_1 \equiv \angle A_2\;\;\;or \;\;\;\measuredangle A_1+\measuredangle A_2=180^{\circ} \tag {$\lambda_A$}$$

Analogously we have

$$\angle B_1 \equiv \angle B _2\;\;\;or \;\;\; \measuredangle B_1+\measuredangle B_2=180^{\circ} \tag{$\lambda_B$};$$

$$\angle C_1 \equiv \angle C_2\;\;\;or \;\;\; \measuredangle C_1+\measuredangle C_2=180^{\circ} . \tag{$\lambda_C$}$$

If in two of the relations $(\lambda_{A,B,C}$ the case of supplementarity would appear, e.g.

$$\measuredangle A_1+\measuredangle A_2=180^{\circ}=\measuredangle B_1+\measuredangle B_2$$

then

$\measuredangle C_1+\measuredangle C_2=180^{\circ}-(\measuredangle A_1+\measuredangle B_1)+180^{\circ}-(\measuredangle A_2+\measuredangle B_2)=(180^{\circ}-\measuredangle A_1-\measuredangle A_2)+(180^{\circ}-\measuredangle B_1-\measuredangle B_2)=0$

impossible. 

          Therefore, in two of the relations $(\lambda_{A,B,C})$ we have $"\equiv "$ (and implicitly, based on the sum of the angles of a triangle we have $"\equiv "$ in the remaining case as well). The triangles will be similar.  

$\square $<end Proof of L_1>

             Lemma 2 Two triangles with respective perpendicular sides are similar.

               Same justification.

                    Proof of (e) [apud E. RUSU, "Cum gindim si rezolvam 200 de probleme", Ed. Albatros, Buc., 1972, pp 226-227, Probl. 176]

              In [the triangles] $\Delta AHB$ and $\Delta A'OB'$ we have

$$AH \perp BC \perp OA'$$

so $AH \parallel AB$; analogous $BH \parallel OB'$. Then $A'B' \parallel AB$ from the middle theorem. 

        With the Lemma 1 result $\Delta AHB \sim \Delta A'OB'$, having the similarity ratio $\frac{AB}{A'B'}=2.$ So also $\frac{AH}{A'O}=2.$

               Remark CiP The fact that (e) also occurs as vectors  (i.e. (E)) results from the observations:

($\alpha$)   $O,\;H\;\in Int \Delta ABC\;\;\Leftrightarrow\;\widehat{BAC},\;\widehat{ABC},\;\widehat {BCA}\;<90^{\circ}$

($\beta$)   $O,\;H\;\in Ext \Delta ABC\;\;\Leftrightarrow\;\widehat{A}\;>90^{\circ}\;or\;\widehat{B}\;>90^{\circ}\;or\; \widehat{C}\;>90^{\circ}$

($\gamma$)  $H=A,\;O=A'\;\;\Leftrightarrow\;\widehat{A}=90^{\circ}.$

Moreover, in the case $(\beta)$ we can specify:

          $\widehat{BAC}>90^{\circ}\;\Rightarrow\;H\in Int \{opposite\;angle\;at\;the\;apex\;with\; \angle BAC\}$

                                                                                   and $O\in \{the\;half-plane\;determined\;by\;the\; line\;BC,\;which\;does\;\underline{not}\;contain\;the\;vertex\;A \}$

$\blacksquare \blacksquare$

Bulgarian Mathematics Magazine for Students

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You can find other Mathematics magazines here.

МАТЕМАТИКА В ШКОЛЕ 1989. 𝒩𝑜 5

 Click on the image to download. The password to open the file is :  ogeometrie  

Other issues of the magazine can be found here.

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