Summer Program COSMOS 2000

R. GARFIELD AND G. ELIAS
GRAND PRIX IN MATH

Irvine, California, July, 2000

Problem 1. Find the remainder of 3²⁰⁰⁰ when divided by 7.

Solution. By Fermat's Little Theorem, we have $3^6\equiv1\,({\mathrm{mod}}\,~7)$. Therefore $3^{1998}\equiv1\,({\mathrm{mod}}\,~7)$and so, $3^{2000}\equiv3^2\,({\mathrm{mod}}\,~7)\equiv2\,({\mathrm{mod}}\,~7)$. The answer is 2.

Problem 2. A natural number n is called ``excellent'' if there exist natural numbers $a_1,a_2\ldots,a_k$ such that:

$$n=a_1+a_2+\cdots+a_k \quad {\mathrm{and}}\quad \frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_k}=1.$$

a)

If n is an excellent number, prove that 2n+2 is excellent as well.

b)

It is known that all numbers between 33 and 73 are excellent. Prove that every natural number $n\ge73$ is also an excellent number.

Solution. a) Assume that n is an excellent number. By definition, $n=a_1+a_2+\cdots+a_k$ for some number $a_1,\dots,a_k$ such that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_k}=1$. Then

$$2n+2=2a_1+2a_2+\cdots+2a_k+2$$

and

$$\frac{1}{2a_1}+\frac{1}{2a_2}+\cdots+\frac{1}{2a_k}+\frac{1}{2}=\frac{1}{2}+\frac{1}{2}=1.$$

b) Notice that 2n+9 is an excellent number. Indeed,

$$2n+9=2a_1+2a_2+\dots+3+6$$

and

$$\frac{1}{2a_1}+\frac{1}{2a_2}+\cdots+\frac{1}{2a_k}+\frac{1}{3}+ \frac{1}{6}=1.$$

Now we apply induction to finish the argument: $74=36\cdot2+2$, $75=33\cdot2+9$, and so forth.

Problem 3. Let ABC be an arbitrary triangle. Prove that there exist 3 congruent circles k₁, k₂ and k₁ passing through one point and such that k₁ is tangent to the sides of the triangle AB and AC, k₂ -- to BC and AB and k₃ -- to AC and BC.

Solution. Denote the centers k₁, k₂ and k₃ (supposing that such exist) by O₁, O₂ and O₃, respectively. Then $\bigtriangleup ABC\sim\bigtriangleup O_1O_2O_3$ by a homothety with center -- the incenter O of $\bigtriangleup ABC$. Obviously, O is also the incenter of $\bigtriangleup O_1O_2O_3$. (See Figure 1)

  

Figure 1:

Let Q be the intersection point of k₁, k₂ and k₃. Observe that Q is the cyclocenter of $\bigtriangleup O_1O_2O_3$; denote by x the radius of the circumscribed circle. (See Figure 2)

For the coefficient c of the similarity of $\bigtriangleup ABC$ and $\bigtriangleup O_1O_2O_3$, we have:

$$c=\frac{r-x}{r}=\frac{x}{R},$$

where r and R are correspondingly the radii of the incircle and circumcircle of $\bigtriangleup ABC$.

Hence,

$$x=\frac{rR}{r+R} \quad {\mathrm {and}}\quad c=\frac{r}{r+R}.$$

We can now construct $\bigtriangleup O_1O_2O_3$ as the image of $\bigtriangleup ABC$ under the dilation with center O (the incenter of $\bigtriangleup ABC$) and coefficient $c=\frac{r}{r+R}$. Denote by R₁ and r₁ the radii of the incircle and circumcircle of $\bigtriangleup O_1O_2O_3$. Then

$$R_1=c\cdot R=\frac{rR}{r+R} \quad {\mathrm {and}}\quad r_1=c\cdot r=\frac{r^2}{r+R}.$$

  

Figure 2:

Hence, r-r₁=R₁, and O₁, O₂ and O₃ are centers of three circles k₁, k₂ and k₃ which exactly meet the requirement of the the problem.

Problem 4. Let k be a circle with center O and l be a line not intersecting k. Drop the perpendicular from O to l and denote by E the intersection point of this perpendicular with l. Let M be an arbitrary point on l, $M\not = E$, let MA and MB be the the two tangents through M (A and B lie on k). Finally, let EC and ED be the perpendiculars from E to MA and MB, respectively. Prove that the intersection point of CD and OE is independent of the choice of M.

Solution. Denote by t₁ and t₂ the tangent lines MA and MB respectively. Denote, also, by l₁, l₂ and l₃ the lines EC, ED and OErespectively. (See Figure 3)

  

Figure 3:

Let h be the power of the point E with respect to k. Consider inversion I with center E and radius $\sqrt{h}$. By the properties of the inversion we have that I(k)=k, I(t₁)=k₁, I(t₂)=k₂, I(l₁)=l₁, I(l₂)=l₂ and I(l₃)=l₃, where k₁ and k₂ are circles passing through Eand tangent to k.The circle with diameter EM passes passes also through D and C (because of the right triangles there). This implies that I(M)=M', I(C)=C' and I(D)=D' are collinear and lie on the line through M' which is perpendicular to l. Notice, also, that k₁ and k₂ pass through M'. Let $k_1\cap l_3=\{C_1', E\}$ and $k_2\cap l_3=\{D_1', E\}$. The image of the line DC under I is then a circle k₃ passing through E, C'and D'.

  

Figure 4:

To prove the problem it has to be shown that the intersecting points of k₃ and l₃ are independent of the choice of M. In other words, if $\{E, E'\}=l_3\cap k_3$, it has to be shown that E' is a constant point on l₃.

The center O₃ of k₃ lies on the line p through the midpoints of C'D' and EE' (because these two segments are parallel chords in k₃). Since D'C' and EE' are perpendicular to l, the line p is parallel to l. The quadrilateral C₁'C'D'D₁' is thus a rectangle and, therefore, p passes through the midpoints of C'D' and C₁'D₁'. The midpoints of EE' and C₁'D₁' then coincide; denote that point by G. Since E' is independent of M if and only if G is independent of M, we will concentrate on the latter. Indeed, we will show that G is the center of k.

  

Figure 5:

Consider $\bigtriangleup C_1'M'D_1''$; G, O₁ and O₂ are midpoints of its sides. Hence |GO₂|=r₁ and |GO₁|=r₂ (the radii of k₁ and k₂). It follows that the circle with center G and radius r₂-r₁ will be tangent to both k₁ and k₂. But there is one only circle with center on C₁'D₁'' which is tangent to k₁ and k₂ and it is k, i.e., G=O. This concludes the proof.