Using knowledge of planimetry for solving some systems of algebraic equations

Aim: Perfecting students' skills for transferring knowledge of planimetry (metric dependencies in a right triangle, triangle's area and cosine rule) for solving systems of quadratic equations with tree or two unknowns (9th and 10th grade).

Example 1. If tema3_EN_files\tema3_EN_MathML_0.jpg and tema3_EN_files\tema3_EN_MathML_1.jpg are positive numbers and tema3_EN_files\tema3_EN_MathML_2.jpg then without finding the values for tema3_EN_files\tema3_EN_MathML_3.jpg and tema3_EN_files\tema3_EN_MathML_4.jpg find the value of the expression tema3_EN_files\tema3_EN_MathML_5.jpg

Solution: If we have to solve this system it would be easy for anyone manage to find tema3_EN_files\tema3_EN_MathML_6.jpg and tema3_EN_files\tema3_EN_MathML_7.jpg but in this problem tema3_EN_files\tema3_EN_MathML_8.jpg is to be found. The system can be written down in the form


Since tema3_EN_files\tema3_EN_MathML_10.jpg and tema3_EN_files\tema3_EN_MathML_11.jpg are positive, from the first equation tema3_EN_files\tema3_EN_MathML_12.jpg and tema3_EN_files\tema3_EN_MathML_13.jpg can be considered using a reverse Pythagorean theorem as the lengths respectively of the legs and the hypotenuse in a right triangle ABD tema3_EN_files\tema3_EN_MathML_14.jpg (fig.1)

Figure 1.

If we consider the second equation of the system we can reach an analogical conclusion tema3_EN_files\tema3_EN_MathML_15.jpg и tema3_EN_files\tema3_EN_MathML_16.jpg are the respective lengths of the legs and hypotenuse in a triangle BCD tema3_EN_files\tema3_EN_MathML_17.jpg(fig.1).

From the third equation we can make the conclusion that y is a number which is a geometrical average of tema3_EN_files\tema3_EN_MathML_18.jpg and tema3_EN_files\tema3_EN_MathML_19.jpg, and using the reverse theorem for proportional line segments in a right triangle the consequence is that tema3_EN_files\tema3_EN_MathML_20.jpg

Now we examine the expression tema3_EN_files\tema3_EN_MathML_21.jpg, which can be represented as tema3_EN_files\tema3_EN_MathML_22.jpg. Considering figure 1 tema3_EN_files\tema3_EN_MathML_23.jpgbut tema3_EN_files\tema3_EN_MathML_24.jpg

In this way we find out that tema3_EN_files\tema3_EN_MathML_25.jpg i.e. tema3_EN_files\tema3_EN_MathML_26.jpg

For this problem the question could be to find tema3_EN_files\tema3_EN_MathML_27.jpg.

Example 2. If tema3_EN_files\tema3_EN_MathML_28.jpg and tema3_EN_files\tema3_EN_MathML_29.jpg are positive numbers and


then without finding values for tema3_EN_files\tema3_EN_MathML_31.jpg and tema3_EN_files\tema3_EN_MathML_32.jpg, calculate the expression tema3_EN_files\tema3_EN_MathML_33.jpg .

Solution: The given system we represent in the form


Since tema3_EN_files\tema3_EN_MathML_35.jpg and tema3_EN_files\tema3_EN_MathML_36.jpg, considering the second equation of the system and having in mind a reverse Pythagorean theorem the numbers tema3_EN_files\tema3_EN_MathML_37.jpg and tema3_EN_files\tema3_EN_MathML_38.jpg are the lengths respectively of the legs and hypotenuse in a right triangle AMC tema3_EN_files\tema3_EN_MathML_39.jpg(fig.2).

If we consider the first equation the numbers tema3_EN_files\tema3_EN_MathML_40.jpg, tema3_EN_files\tema3_EN_MathML_41.jpgand tema3_EN_files\tema3_EN_MathML_42.jpg are lengths of sides in a triangle AMB in whichtema3_EN_files\tema3_EN_MathML_43.jpg and using a reverse cosine rule. Analogically tema3_EN_files\tema3_EN_MathML_44.jpg, tema3_EN_files\tema3_EN_MathML_45.jpgand tema3_EN_files\tema3_EN_MathML_46.jpg are lengths of sides in a triangle BMC where tema3_EN_files\tema3_EN_MathML_47.jpg ( fig. 2).

Figure 2.

Since tema3_EN_files\tema3_EN_MathML_48.jpg, then using a reverse Pythagorean theorem we can conclude that ABC is a right triangle and tema3_EN_files\tema3_EN_MathML_49.jpg ACB = tema3_EN_files\tema3_EN_MathML_50.jpg

We will find the areas of the following three triangles: tema3_EN_files\tema3_EN_MathML_51.jpg using which the area of tema3_EN_files\tema3_EN_MathML_52.jpgABC can be found.


This way we get tema3_EN_files\tema3_EN_MathML_57.jpgtema3_EN_files\tema3_EN_MathML_58.jpg, i.e. tema3_EN_files\tema3_EN_MathML_59.jpg .

For individual work

Problem 1. If tema3_EN_files\tema3_EN_MathML_60.jpgand tema3_EN_files\tema3_EN_MathML_61.jpg are positive numbers and tema3_EN_files\tema3_EN_MathML_62.jpg then without finding the values of tema3_EN_files\tema3_EN_MathML_63.jpg and tema3_EN_files\tema3_EN_MathML_64.jpg find the value of the expression tema3_EN_files\tema3_EN_MathML_65.jpg

Answer: 12.

Problem 2. Find tema3_EN_files\tema3_EN_MathML_66.jpg, if tema3_EN_files\tema3_EN_MathML_67.jpg, tema3_EN_files\tema3_EN_MathML_68.jpg, tema3_EN_files\tema3_EN_MathML_69.jpg and


Answer: tema3_EN_files\tema3_EN_MathML_71.jpg.

Problem 3. If tema3_EN_files\tema3_EN_MathML_72.jpg, tema3_EN_files\tema3_EN_MathML_73.jpg, tema3_EN_files\tema3_EN_MathML_74.jpg does the system have a solution:


Hint: If tema3_EN_files\tema3_EN_MathML_76.jpg are solutions to the system, than the geometric interpretation of the system is presented in figure 3. Use the inequality between the sides of a triangle i.e. tema3_EN_files\tema3_EN_MathML_77.jpg

Figure 3.

Answer: No.

Problem 4. Calculate the value of the expression tema3_EN_files\tema3_EN_MathML_78.jpg, if tema3_EN_files\tema3_EN_MathML_79.jpg and


Hint: tema3_EN_files\tema3_EN_MathML_81.jpg and tema3_EN_files\tema3_EN_MathML_82.jpg, because if tema3_EN_files\tema3_EN_MathML_83.jpg and tema3_EN_files\tema3_EN_MathML_84.jpg, then tema3_EN_files\tema3_EN_MathML_85.jpg. The given system we represent in the form


Look above at the solution of problem 1.

Problem 5. Solve the system tema3_EN_files\tema3_EN_MathML_87.jpg .

Hint: It is not hard to prove that tema3_EN_files\tema3_EN_MathML_88.jpg and tema3_EN_files\tema3_EN_MathML_89.jpg are positive numbers and tema3_EN_files\tema3_EN_MathML_90.jpgand tema3_EN_files\tema3_EN_MathML_91.jpgare lengths of the legs and hypotenuse in a right triangle.

Answer: tema3_EN_files\tema3_EN_MathML_92.jpg.

By Rumyana Mavrova, Plovdiv university,