## Solutions for exercises "C" in May, 2002 |

In this page only the sketch of the solutions are published; in some cases only the final results. To achieve the maximum score in the competition more detailed solutions needed.

**C. 675.** Is there a square number in
which the last two digits are both odd?

**Solution.** The number in question can only be the square of an odd number, of the form (2*k*+1)^{2}=4*k*(*k*+1)+1, thus it leaves a remainder of 1 when divided by 4 (or even by 8). The last digit of an odd square number may be 1, 5 or 9, which now means that the remainder is 11, 15 or 19 when the number is divided by 20. Such numbers, however leave a remainder of 3 when divided by 4, and that contradicts to the observation above. Therefore, the last two digits of a square number cannot both be odd.

**C. 676.** The length of side *AB* of
the rectangle *ABEF* is 1 unit, and the length of side
*BE* is 3 units. The points *C* and *D* divide
side *BE* into three equal parts. Show that *BAC*+*BAD*\(\displaystyle \angle\)+*BAE*=180^{o}.

**Solution.**

The figure shows that *BAD*'\(\displaystyle mangle\)=*BAD*\(\displaystyle mangle\) and *BAC*\(\displaystyle mangle\)=45^{o}=*EAA*'\(\displaystyle mangle\), thus *BAD*\(\displaystyle mangle\)+*BAE*\(\displaystyle mangle\)+*BAC*\(\displaystyle mangle\)=*D*'*AB*\(\displaystyle mangle\)+*BAE*\(\displaystyle mangle\)+*EAA*'=180^{o}. (The problem can also be solved by using the addition formula of the tangent function.)

**C. 677.** Find the integers *a* and
*b*, given that
*a*^{4}+(*a*+*b*)^{4}+*b*^{4}
is a square number.

**Solution.** A square number may leave a remainder of 0 or 1 when divided by 4. 0 if the number is even and 1 if it is odd. Thus there is at most one odd number among *a*, *b*, and *a*+*b*. That can only happen if *a* and *b* are both even: *a*=2*a*_{1}, *b*=2*b*_{1}. Hence if *a*^{4}+(*a*+*b*)^{4}+*b*^{4}=16(*a*_{1}^{4}+(*a*_{1}+*b*_{1})^{4}+*b*_{1}^{4}) is a perfect square, then so is *a*_{1}^{4}+(*a*_{1}+*b*_{1})^{4}+*b*_{1}^{4}. That makes *a*_{1} and *b*_{1} also even, and so on. It follows that *a* and *b* are divisible by every power of 2, which is only possible if *a*=*b*=0; and that satisfies the requirement of the problem, too.

**C. 678.** In a triangle *ABC*,
*AC*=1, *ABC*\(\displaystyle \angle\)=30^{o}, *BAC*=60^{o}, and *D* denotes the foot of
the altitude from vertex *C*. Find the distance between the
centres of the inscribed circles of the triangles *ACD* and
*BCD*.

**Solution.**

*ABC*

is a (special) right-angled triangle, \(\displaystyle BC=\sqrt{3}\), *AB*=2^{.}*AC*=2. The perimeter of the triangle *ACD* is \(\displaystyle k_{ACD}={1\over2}k_{ABC}={{3+\sqrt{3}}\over2}\), and similarly, \(\displaystyle k_{BCD}={{\sqrt{3}}\over2}k_{ABC}={{3+3\sqrt{3}}\over2}\). The corresponding areas are \(\displaystyle t_{ACD}={1\over4}t_{ABC}={1\over4}\cdot{{\sqrt{3}}\over2}={{\sqrt{3}}\over8}\), \(\displaystyle t_{BCD}={3\over4}t_{ABC}={3\over4}\cdot{{\sqrt{3}}\over2}={{3\sqrt{3}}\over8}\). Thus the radii of the incircles of the small triangles are \(\displaystyle KE={{2\cdot t_{ACD}}\over{k_{ACD}}}={{\sqrt{3}}\over4}/{{3+\sqrt{3}}\over2}={{\sqrt{3}-1}\over4}\), \(\displaystyle LF={{2\cdot t_{BCD}}\over{k_{BCD}}}={{3\sqrt{3}}\over4}/{{3+3\sqrt{3}}\over2}={{3-\sqrt{3}}\over4}\). Finally, \(\displaystyle ED={{k_{ACD}}\over2}-AC={{3+\sqrt{3}}\over4}-1={{\sqrt{3}-1}\over4}\), \(\displaystyle DF={{k_{BCD}}\over2}-BC={{3+3\sqrt{3}}\over4}-\sqrt{3}={{3-\sqrt{3}}\over4}\), thus it follows from the Pythagorean theorem that \(\displaystyle KL=\sqrt{EF^2+(LF-KE)^2}=\sqrt{{1\over4}+{{(2-\sqrt{3})^2}\over4}}={2-\sqrt{3}}\approx0.52\).

**C. 679.** Given three spheres that
pairwise touch each other and also touch the plane *S*, determine
the radius of the sphere that has its centre on the plane *S*,
and that touches all the three spheres.

**Solution.**

The centres of the three unit spheres form an equilateral triangle of side length 2 in a plane parallel to the plane *S* at a distance of 1. The distance of the centre *K* of the sphere of radius *r* in question from all three vertices of the triangle is *r\(\displaystyle pm\)*1, depending on whether it touches the spheres on the outside or on the inside. Let *C* denote one of the three centres, and let *C*' be its orthogonal projection onto the plane *S*. It follows from the Pythagorean theorem applied to the right triangle *KC*'*C* that \(\displaystyle KC'=\sqrt{KC^2-C'C^2}=\sqrt{(r\pm1)^2-1}=\sqrt{r^2\pm2r}\). On the other hand, the length of *KC*' is equal to the distance \(\displaystyle {2\over{\sqrt{3}}}\) of the centre of the equilateral triangle of 2-unit side from the vertices. Hence \(\displaystyle r^2\pm2r={4\over3}\), thus \(\displaystyle r={{\sqrt{21}\pm3}\over3}\). So far, we have been assuming that the sphere in question either touches all the unit spheres on the outside or it touches all of them on the inside. There is no "mixed" configuration, because if the sphere of radius *r* of centre *K* touched the sphere of centre*G*_{1} on the outside and the sphere of centre *G*_{2} on the inside, then *KG*_{1}=*r*+1, *KG*_{2}=*r*-1 would imply that *G*_{1}*G*_{2}=2=(*r*+1)-(*r*-1)=*KG*_{1}-*KG*_{2}, that is, the triangle *G*_{1}*G*_{2}*K* would be degenerate, which means that *K* would lie on the line *G*_{1}*G*_{2}. That line, however, is parallel to the plane *S*, and thus has no common point with it.