p1. Show that if $a, b, c$ are odd integers then the equation $ax^2 + bx + c = 0$ has no rational roots. p2. If $k$ is a positive integer, find the greatest power of $3$ that divides $10^k-1$. p3. The figure shows the triangle $ABC$, right at $C$, its circumscribed circle and semicircles built on the two legs. Show that the sum of the areas of the two shaded regions is $\frac12 AC\cdot CB$. p4. Each vertex of a cube is assigned the value $+1$ or $-1$, and each face the product of the values assigned to its vertices. What values can the sum of the $14$ numbers thus obtained, have? p5. Consider the regular pentagon $ABCDE$ in the figure. If $BI$ is $ 1$, how long is $AB$? p6. Let $n\ge 3$ be an integer. A circle is divided into $2n$ arcs by $2n$ points. Each arch measures one of three possible lengths, and no two adjacent arches are the same length. The $2n$ points are alternately colored red and blue. Show that the $n$-gon with blue vertices and the $n$-gon with red vertices have the same perimeter and the same area. PS. Seniors P3,P4 were also posted as Juniors P1, P5.
Problem
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Tags: algebra, geometry, combinatorics, number theory, chilean NMO
FIREDRAGONMATH16
14.10.2021 23:32
We factor $10^k-1$ as follows:
$10^k-1=(10-9)(10^{k-1}+10^{k-2}+\cdots+10^2+10+1)=9(\underbrace{111\cdots11}_\text{k 1's})$
Now we clearly see that the greatest power of $3$ is $3^{2+v_3{k}}$ .
[asy][asy]
draw(circle((0,0),5));
draw((-5,0)--(3,4)--(5,0)--cycle);
draw(arc((-1,2),(3,4),(-5,0)));
draw(arc((4,2), (5,0), (3,4)));
label((5.2, 2.6), "B");
label((-2,5.5), "A");
label((-1.5, 3.2),"C");
label((2,2),"D");
label((4.25,2.25), "E");
[/asy][/asy]
Label the areas as $A,B,C,D$ and $E$ as shown in the diagram. Let each leg have lengths $x$ and $y$ respectively. Then, the hypothenuse will have length $2\sqrt{x^2+y^2}$ and thus the radius will be $\sqrt{x^2+y^2}$. The area of the semicircle is
$$\frac{1}{2}(\sqrt{x^2+y^2})^2 \pi = \frac{1}{2}(x^2\pi + y^2\pi)$$
Also, the radius of each of the semicircles are $x$ and $y$ respectively. Now with a bit of careful manipulation:
$$\frac{1}{2}x^2\pi+\frac{1}{2}y^2\pi = \frac{1}{2}x^2\pi+\frac{1}{2}y^2\pi$$
We can express these two areas in terms of our variables!!
$$C+D+E = (A+C)+(B+E)$$
$$\boxed{A+B=D}$$
as desired.
ah screw this asymptote diagram @below yea that seems correct...
jasperE3
15.10.2021 00:09
$v_3(10^k-1)=v_3(10-1)+v_3(k)=2+v_3(k)$ by LTE, so the answer is $3^{2+v_3(k)}$.
iniffur
15.10.2021 02:17
Solution 1 For having rational roots the discriminant $b^2-4ac$ should be a perfect square. In the present case, where a, b and c are all odd, it would be an odd square, which taken mod 8 should give 1, contrary to $b^2-4ac$ which actually returns -3 modulo 8 $(b^2\equiv1$ mod 8 and $4ac\equiv 4$ mod 8). Contradiction.