ogeometrie

joi, 20 ianuarie 2022

Problem E:2718, not from AMERICAN MATHEMATICAL MONTHLY

It is from the Romanian magazine "Gazeta Matematica"

Proposed for 7th grade on page 224. The statement is (thanks to Miss Google for the translation):

"Let the polynomial be given $P(x,y)=by(x^2+a^2)-ax(y^2+b^2)+xy.$ Show that:

$$P(b,a)=P(-\frac{a}{4},-4b)=P(\frac{b}{a^2-b^2},\frac{a}{a^2-b^2})=P(-\frac{5ab+1}{5b},\frac{4-5ab}{5a})."$$

Solution CiP

It's easy to calculate $P(b,a)=b \cdot a(a^2+b^2)-a \cdot b(a^2+b^2)+b \cdot a=a \cdot b.$

To avoid laborious calculations, we will write the polynomial in a different way.

$$P(x,y)=\underline{bx^2y}+\underline{\underline{a^2by}}\; \underline{-axy^2}\;\underline{\underline{-ab^2x}}+xy,$$

now we group the identically-underlined terms,

$$P(x,y)=xy(bx-ay)+ab(ay-bx)+xy,$$

$$P(x,y)=(xy-ab)\cdot (bx-ay)+xy. \tag{1}$$

For the couple $(x,y)=(-\frac{a}{4},-4b)$ we see that $xy=ab$ so the first paranthesis in the formula $(1)$ is equal to zero, finally obtaining the result $P(x,y)=ab.$

For the couple $(x,y)=(\frac{b}{a^2-b^2}, \frac{a}{a^2-b^2})$ we see that

$$bx-ay=\frac{b^2}{a^2-b^2}-\frac{a^2}{a^2-b^2}=-1.$$

So for these two values of $x$ and $y$ we have from $(1)$ that

$$P(x,y)=(xy-ab)(-1)+xy=ab.$$
The same goes for the couple $(x,y)=(-\frac{5ab+1}{5b},\frac{4-5ab}{5a})$; here again

$$bx-ay=-\frac{5ab+1}{5}-\frac{4-5ab}{5}=\frac{-5ab-1-4+5ab}{5}=-1.$$

All values we calculated are equal to $ab$.

$\blacksquare$

miercuri, 19 ianuarie 2022

G M B #5 - 1967 (contine articolul ARIA TRIUNGHIULUI IN FUNCTIE DE MEDIANE - pag. 205-206)

Vezi in DRIVE

Vezi ERATA (necorectat inca)

Articolul de la pp. 205-206 l-am prezentat in anul 1976 la Cercul de Matematica al scolii (Lic Energetic Timisoara), sub conducerea prof. Vasile BIVOLARU....

Am "generalizat" Aplicatia prezentata la sfarsit, aratand ca:

Daca medianele unui triunghi verifica relatia $m_a^2 + m_b^2 =m_c^2$ atunci aria triunghiului este egala cu

$A_{\Delta ABC}=\frac{2}{3}\cdot m_a \cdot m_b$.

Nu mai stiu cum am demonstrat; ar merge cam asa:

Avem formula generala

$$A_{\Delta ABC}=\frac{1}{3}\cdot \sqrt{2 \cdot \sum m_a^2 \cdot m_b^2 - \sum m_a^4}.$$

Calculam expresia de sub radical, inlocuind $m_c^2$ cu $m_a^2+m_b^2$ si obtinem

$2 \cdot (m_a^2 \cdot m_b^2+(m_a^2+m_b^2) \cdot m_c^2))-m_a^4-m_b^4-(m_a^2+m_b^2)^2=$

$=2(m_a^2 m_b^2 +(m_a^2 +m_b^2)^2)-m_a^4-m_b^4-(m_a^2+m_b^2)^2=2m_a^2 m_b^2+(m_a^2+m_b^2)^2-m_a^4-m_b^4=4m_a^2 m_b^2$

Deci $A_{\Delta ABC}=\frac{1}{3}\cdot \sqrt{4m_a^2m_b^2}=\frac{2}{3}m_a m_b$.

$\blacksquare$