Elementary creation of flat geometrical figures, such, as circles and triangles which can surprise fans of mathematics.
Instruction
1. Of course in our modern century such elementary figures on the plane as a triangle and a circle it is difficult to surprise someone. They are studied long ago, the laws allowing to calculate all their parameters are output long ago. But sometimes at the solution of various tasks it is possible to face surprising things. Let's consider one interesting construction. Let's take any triangle of AVS which has a party the EXPERT – big of the parties, and we will do the following:
2. At first we build a circle with center "A" and radius equal to the party of a triangle of "AV". The EXPERT we will designate a circle point of intersection with the party of a triangle as a point of "D".
3. Then we cost a circle with center "C" and radius equal to a piece of "CD". We will designate a point of intersection of the second circle with the party of a triangle of "SV" as the point "Е".
4. We build the following circle with center "B" and radius equal to a piece of "VE". We will designate a point of intersection of thirds of a circle with the party of a triangle of "AV" as a point of "F".
5. We build the fourth circle with center "A" and radius equal to a piece of "AF". "EXPERT" we will designate a point of intersection of the fourth circle with the party of a triangle as a point "To".
6. And we build the last, fifth circle with center "C" and radius of "SK". In this construction interestingly following: the triangle top "In" accurately gets on the fifth circle.
7. For fidelity it is possible to try to repeat construction using a triangle with other lengths of the parties and corners with only one condition that the party "EXPERT" the greatest of the parties of a triangle, and all the same the fifth circle will accurately get to top "In". It means only one: it has radius equal to the party of "SV", respectively the piece of "SK" is equal to the party of a triangle of "SV".
8. The simple mathematical analysis of the described construction, looks as follows. The piece of "AD" is equal to the party of a triangle of "AV" since points "In" and "D" are on one circle. Radius of the first circle of R1 = AV. Piece of CD=AC-AB, that is radius of the second circle: R2=AC-AB. The piece of "SE" is respectively equal to the radius of the second circle of R2, the piece of BE=BC-(AC-AB) means, the radius of thirds of a circle of R3=AB+ VS-ASOtrezok means "BF" is equal to the radius of thirds of a circle of R3, from here a piece of AF=AB-(AB+BC-AC) = AC-BC, that is radius of the fourth circle of R4=AC-BC. The piece of "joint stock company" is equal to the radius of the fourth circle of R4, a piece of CK=AC-(AC-BC)=BC, that is radius of the fifth circle of R5=BC from here.
9. From the received analysis it is possible to draw an unambiguous conclusion that at similar creation of circles with the centers in triangle tops the fifth creation of a circle gives the circle radius equal to the party of a triangle of "VS".
10. Let's continue further reasonings concerning this construction and we will define what the sum of radiuses of circles, and here is equal to that we will receive: ∑R=R1+R2+R3+R4+R5 == AV + (AC-AB)+ (AB+BC-AC)+ (AC-BC) + VS. If to remove the brackets and to bring similar composed, then we will receive the following: ∑R= AB+BC+ ASOchevidno, the sum of radiuses of the received five circles with the centers in triangle tops, is equal to perimeter of this triangle. Also the following is remarkable: pieces of "VE", "BF" and "KD" are equal among themselves and equal to the radius of thirds of a circle of R3. BE=BF=KD=R3=AB+BC-AC
11. Of course, all this is related to elementary mathematics, but, perhaps, has some applied value and can be the cause for further researches.