ogeometrie

miercuri, 8 iunie 2022

A NECESSARY_ and_SUFFICIENT condition of tangency to the circle

Given a triangle $ABC$,

a necessary and sufficient condition for the line $CD$ to be tangent

to the circle circumscribing the triangle is $\angle{BAC} \equiv \angle{BCD}.$

The condition is necessary : $CD$ - is tangent to circle $\Rightarrow\;\angle{BAC}=\angle{BCD}.$

Indeed, we have the equalities between the measures of angles and arcs of the circle:

$$\widehat{BAC}=\frac{\overset{\frown}{BC}}{2},\;\;\widehat{BCD}=\frac{\overset{\frown}{BC}}{2}$$
hence the conclusion.

The condition is sufficient : $\angle{BAC}=\angle{BCD}\;\Rightarrow$ $CD$ - is tangent to the

circle circumscribing the triangle.

For demonstration we take $O$ the center of the circumscribed circle.

If the point $O$ is inside the triangle, then we write the relation

$$\widehat{AOB}+\widehat{BOC}+\widehat{COA}=180^{\circ}.$$

We show the proof when point $O$ is outside the triangle. In this case we have the relationship

$$\widehat{BOA}+\widehat{AOC}=\widehat{BOC}.$$

From the isosceles triangles $AOB\;(OA=OB)$ and $AOC\;(OA=OC)$ it follows

$$\widehat{BOA}=180^{\circ}-2 \cdot \widehat{BAO},\;\;\widehat{AOC}=180^{\circ}-2 \cdot \widehat{CAO}$$

so $\widehat{BOC}=360^{\circ}-2 \cdot (\widehat{BAO}+\widehat{CAO})=360^{\circ}-2 \cdot \widehat{BAC}.$

But in the isosceles triangle $BOC$ we have

$$\widehat{BCO}=\frac{180^{\circ}-\widehat{BOC}}{2}=\frac{2\cdot \widehat{BAC}-180^{\circ}}{2}=\widehat{BAC}-90^{\circ}.$$

Using the hypothesis $\widehat{BAC}=\widehat{BCD}$ results $$\widehat{OCD}=\widehat{BCD}-\widehat{BCO}=\widehat{BAC}-(\widehat{BAC}-90^{\circ})=90^{\circ},$$

so the line $CD$ is perpendicular to the radius $OC$ of the circle circumscribing the triangle $ABC$, i.e. it is tangent to this circle.

$\blacksquare$

As an application of this property we solve the following problem:

(from the GMB exercise supplement, from October 2009 (page 3; proposed for 7th grade at a student summer camp). In translation, thanks to Mr DeepL

"The median $AM$ of the acute triangle $ABC$ intersects the circle

circumscribing the triangle a second time at point $D$. If $E$ is the symmetric

point of $A$ with respect to $M$, then show that the line $BC$ is the common

tangent of the circumscribed circles of triangles $BDE$ and $CDE$."

For demonstration, let's take a simplified figure.

From $MB=MC$ and $MA=ME\;\;\Rightarrow\;BACE$ - parallelogram. We have equals angles:

$$\angle{CAE} \equiv \angle{AEB}\;\;\;\angle{BAE}\equiv \angle {AEC},$$

or written

$$\widehat{CAD}=\widehat{DEB},\;\;\;\widehat{BAD}=\widehat{DEC}.\tag{1}$$

We still have the equalities between angles and arcs:

$$\widehat{CAD}=\frac{\overset{\frown}{CD}}{2}=\widehat{CBD,}\;\;\;\widehat{BAD}=\frac{\overset{\frown}{BD}}{2}=\widehat{BCD}. \tag{2}$$

From (1) and (2) follows

$\widehat{DBC}=\widehat{DEB}$ hence $BC$ is tangent to the circle circumscribed of $\Delta BDE,$

$\widehat{DCB}=\widehat{DEC}$ hence $CB$ is tangent to the circle circumscribed of $\Delta CDE$.

$\blacksquare\;\blacksquare$

SUPLIMENTUL cu EXERCIȚII al GMB N0 10/2009

Vezi in DRIVE

sâmbătă, 4 iunie 2022

Opublikowany problem matematyczny... za wcześnie // A published math problem ... too early

This is problem S:E22.89 from the GMB exercise supplement, from March 2022 (page 5; proposed for 5th grade).

Translated (thanks to DeepL):

"Let the natural numbers $\;a,\;b,\;c,\;x,\;y,\;z,\;$ be greater than $1$ such that

$$a^b\;=\;c,\;\;c^x\;=\;y,\;\;y^z\;=\;a^{2025}.$$

a) Calculate the multiplication $\;b\cdot x\cdot z\;.$

b) Give an example of numbers that check all the above conditions."

It is customary in the current year, for many problems, to contain the number 2022, as a kind of parameter, that can be in the problem any number. The author, Mr MUNTEANU Emanuel George, rushed three years earlier to publish this problem. Or is it "because" of the elegant decomposition of the number 2025 into prime factors : $2025=3^4 \cdot 5^2$ ?

Answer CiP

a) $b \cdot x\cdot z\;=\;2025;$

b) $a=7,\;b=15,\;c=7^{15},\;x=3,\;y=7^{45},\;z=45.$

Indeed, $a^b=7^{15}=c,\;c^x=(7^{15})^3=7^{45}=y,\;y^z=(7^{45})^{45}=7^{45\cdot 45}=7^{2025}=a^{2025}.$

A more general example is of the form:

$a\geqslant 2\;$ fixed natural number;

$z=3^{\alpha}\cdot 5^{\beta},\;\;0 \leqslant \alpha \leqslant 4,\;0 \leqslant \beta \leqslant 2$ but $(\alpha,\beta) \neq (0,0),\;(4,2)$ ; a divisor of the number 2025 ;

$b=3^{\gamma}\cdot 5^{\delta},\;\;0 \leqslant \gamma \leqslant 4-\alpha,\;0 \leqslant \delta \leqslant 2- \beta,$ but not to have both $\gamma =4-\alpha, \delta =2-\beta$; a divisor of the number $\frac{2025}{z};$

$x=3^{4-\alpha-\gamma} \cdot 5^{2-\beta-\delta};$

$y=a^{3^{4-\alpha} \cdot 5^{2-\beta}}.$

Solution CiP

a) $a^{2025}\underset{def}{=}\;y^z \;\underset{y=c^x}{=}\;\left ( c^x \right )^z=c^{x \cdot z}\;\underset{c=a^b}{=}\left ( a^b \right )^{x \cdot z}=a^{b \cdot x \cdot z}$, and from this it follows

$$b \cdot x \cdot z\;=\;2025. \tag{1}$$

b) We see from (1) that $b,x,z$ are divisors of 2025. We must have
$$z=3^{\alpha} \cdot 5^{\beta} \;,\tag{2}$$

with $(\alpha,\beta) \neq (0,0)$ since $z \neq 1$, and $(\alpha,\beta) \neq (4,2)$ since $b,x > 1$.

Beacause $b\cdot x=\frac{2025}{z}=3^{4-\alpha}\cdot 5^{2-\beta}$, we must have

$$b=3^{\gamma}\cdot 5^{\delta}$$

and we choose $\gamma,\; \delta$ so that $b$ and $x$ are greater than 1.

The base $a$ can be any natural number $\geqslant 2$.

In the concrete example we chose $a=7,\; \alpha =2,\; \beta=1,\;\gamma=\delta=1$.

We could count all possible cases for a given $a$, but we won't do it here.

$\blacksquare$