168. Two pedestrians set out towards each other at the same time: the first from point $A$, the second from point $B$. The first pedestrian walked 1 km more before the meeting than the second. The first pedestrian arrived at point $B$ 45 minutes after the meeting. The second pedestrian arrived at point $A$ 1 hour and 20 minutes after the meeting. Find the distance from $A$ to $B$.
S o l u t i o n. Let the speed of the first pedestrian who left point $A$ be $v_{1}$ kilometers per hour, and the speed of the second pedestrian be $v_{2}$ kilometers per hour. Since after meeting, the first pedestrian walked $\frac{3}{4} v_{1}$ kilometers, and the second $-\frac{4}{3} v_{2}$ kilometers, and the first pedestrian walked 1 km less than the second, we have the equation

$$
\frac{4}{3} v_{2}-\frac{3}{4} v_{1}=1
$$

Since before the meeting, the pedestrians spent the same amount of time, we have

$$
\frac{\frac{4}{3} v_{2}}{v_{1}}=\frac{\frac{3}{4} v_{1}}{v_{2}}
$$

Thus, we have the system of equations:

$$
\left\{\begin{array} { l } 
{ \frac { 4 } { 3 } v _ { 2 } - \frac { 3 } { 4 } v _ { 1 } = 1 , } \\
{ \frac { \frac { 4 } { 3 } v _ { 2 } } { v _ { 1 } } = \frac { \frac { 3 } { 4 } v _ { 1 } } { v _ { 2 } } , }
\end{array} \text { or } \quad \left\{\begin{array}{l}
v_{2}=\frac{12+9 v_{1}}{16}, \\
\frac{4}{3} v_{2}^{2}=\frac{3}{4} v_{1}^{2},
\end{array}\right.\right.
$$

from which $7 v_{1}^{2}-24 v_{1}-16=0, v_{1}=4$. Therefore, the distance from $A$ to $B$ is 7 km $\left(2 \cdot 4 \cdot \frac{3}{4}+1=7\right)$.

A n s w e r: 7 km.

## CONTENTS

Preface . . . . . . . . . . . . . . . . . 3

§1. Why do we solve problems in school (on the functions of problems in mathematics education) . . . . . . . . . . 5

§ 2. On the methodology of teaching students to solve non-standard algebraic problems . . . . . . . . . . 11

§3. On the role of observations and induction in finding ways to solve non-standard algebraic problems . . . . . . . . . 19

§4. On finding different ways to solve problems . . . . 24

§ 5. On constructing graphs of functions and dependencies containing the modulus sign . . . . . . . . . . 30

§6. Methodological recommendations for using challenging problems in the process of teaching algebra . . . . . . . . . 35

§ 7. Challenging problems in the algebra course for 7th grade . . . 40

§8. Challenging problems in the algebra course for 8th grade . . . 99

§9. Challenging problems in the algebra course for 9th grade . . . 157

## Educational Publication  Kostrikina Nina Petrovna CHALLENGING PROBLEMS  IN THE ALGEBRA COURSE $7-9$ grades

Editorial Board Chair $T$. A. Burmistrova Editor L. M. Kotoeva

Assistant Editor L. I. Zasedateleva Artists L. B. Nikolaev, V. V. Kostin Art Editor Yu. V. Pakhomov Technical Editor N. T. Rudnikova

Proofreader M. Yu. Sergeeva

IB № 12646

Submitted for typesetting 25. 12. 90. Signed for printing from transparencies 12. 08. 91. Format $60 \times 90^{1 / 16}$. Offset paper. Font. Offset printing. Usual printing sheets $15+0.25$ f. Usual printed sheets 15.75. Usual publication sheets $13.34+0.34$ f. Price 1 r. 90 k.

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## SOME "REMARKABLE"

$$
\begin{aligned}
& (1+3+5+\ldots+(2 n-1))^{2}=n^{2} \\
& 1^{3}+2^{3}+3^{3}+\ldots+n^{3}=(1+2+3+\ldots+n)^{2} \\
& \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{(n-1) n}=\frac{n-1}{n} \\
& \frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}
\end{aligned}
$$

## EXAMPLES OF RULES

1) $\overline{a b} \cdot 11=100 a+10(a+b)+b$

$36 \cdot 11=100 \cdot 3+10(3+6)+b=391$

$\overline{a b c} \cdot 11=1000 a+100(a+b)+10(b+c)+c$

$243 \cdot 11=1000 \cdot 2+100(2+4)+10(4+3)+3=$
2) $\overline{a 5}^{2}=a(a+1) \cdot 100+25$

$65^{2}=6 \cdot 7 \cdot 100+25=4225$

3) If $b+c=10$, then $\overline{a b} \cdot \overline{a c}=100 a(a+1)+b c$ $38 \cdot 32=3 \cdot 4 \cdot 100+8 \cdot 2=1216$

## INEQUALITIES

$$
\begin{aligned}
& a+\frac{1}{a} \geqslant 2 \text { for } a>0 \\
& \frac{a+b}{2} \geqslant \sqrt{a b}, \frac{a+b+c}{3} \geqslant \sqrt{a b c} \text { for } a \geqslant 0, b \geqslant 0, c \geqslant 0 \\
& (a+b)(b+c)(c+a) \geqslant 8 a b c \quad \text { for } a>0, b>0, c>0 \\
& \sqrt{(a+c)(b+d)} \geqslant \sqrt{a b}+\sqrt{c d} \text { for } a \geqslant 0, b \geqslant 0, c \geqslant 0, d \geqslant 0
\end{aligned}
$$

## MENTAL CALCULATION

$$
\begin{aligned}
& \text { 4) }(a+b)(a-b)=a^{2}-b^{2} \\
& 71 \cdot 69=(70+1)(70-1)= \\
& =4900-1=4899 \\
& 111 \cdot 89=(100+11)(100-11)= \\
& =10000-121=9879 \\
& \text { 5) } a^{2}=a^{2}-b^{2}+b^{2}=(a+b)(a-b)+b^{2} \\
& 27^{2}=(27+3)(27-3)+3^{2}=729
\end{aligned}
$$


[^0]:    ' Pólya G. Mathematical Discovery - M. Science, $1970-$ P. 16

[^1]:    ' Problems $1-47$ in parentheses indicate the exercise numbers in the textbook $|4|$ 40

[^2]:    ' Problems $48-79$ in parentheses indicate the exercise numbers in the textbook |I| 60

[^3]:    ' Problems $80-120$ in parentheses indicate the exercise numbers in the textbook [7].

[^4]:    ' Problems $1-64$ in parentheses indicate the exercise numbers in the textbook [6].

[^5]:    ${ }^{1}$ Problems $65-107$ in parentheses indicate the exercise numbers in the textbook [3].