Spain | A Nice Algebra Problem | Math Olympiad
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- čas přidán 7. 07. 2024
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(abc)² = 100 * 200 * 300 = 6000000
abc = ±1000√6
a + b + c = abc/bc + abc/ac + abc/ab
= abc(1/200 + 1/300 + 1/100)
= abc/100(1/2 + 1/3 + 1)
= ±(55√6)/3
@yuki2go дал ЭЛЕГАНТНОЕ аналитическое решение, которое правда не дает индивидуальных значений a, b и c. Чего, впрочем и не требовалось. Для их нахождения потребовались бы дополнительные шаги.
Позволю объяснить решение по шагам, надеюсь будет полезно школьникам, подобным мне.
1. (abc)² = 100 * 200 * 300 = 6000000
Это следует из перемножения всех трех уравнений: (ab)(bc)(ca) = 100 * 200 * 300
2. abc = ±1000√6
Это квадратный корень из 6000000. Знак ± появляется, так как квадратный корень может быть положительным или отрицательным.
3. a + b + c = abc/bc + abc/ac + abc/ab
Это ключевой шаг. Мы выражаем a, b и c через их произведение и попарные произведения.
4. = abc(1/200 + 1/300 + 1/100)
Здесь мы заменяем bc, ac и ab их значениями из исходных уравнений.
5. = abc/100(1/2 + 1/3 + 1)
Упрощаем выражение, вынося общий множитель.
6. = ±(55√6)/3
Подставляем значение abc = ±1000√6 и упрощаем:
±1000√6/100 * (1/2 + 1/3 + 1) = ±10√6 * (11/6) = ±(55√6)/3
@@Yuri_Kravets лишние действия в ответе.
±110/√6
Таки красивше))
It's a nice problem, but I guess you could have solved it faster if you had first isolated b , this way: b = 100/a; b = 200/c; c = 2a. I found a = 5 sqrt 6; c = 10 sqrt 6; b = (10/3) sqrt 6; a+b+c = (55/3) sqrt 6; but I did not have to deal with big numbers, like root of 6050. I wonder if my solution has something wrong, but I think it's easier. Anyway, thanks for the problem.
I solved the problem in the same way. Final result: +- 110/sqrt6 giving the same result as +- 55 sqrt6/3.
agree with
just see i know that
c^2=300*2
c=10sqare6
so dont life to hard
absolutely agree 👍🏻
I solved the same way
AB=100
B=100/A so from BC=200 we have 100C/A=200 ie C=2A
Then from CA=300 we have 2A^2=300 ie A=sqrt(150)
We can deduce that:
C=2sqrt(150) since C=2A and B=200/(2sqrt(150))=100sqrt(150)/150
After 3:14 (the value of a^2, b^2 and c^2), you can continue by :
ab > 0, bc > 0 and ac > 0 so a, b and c are positives.
And now you know the value of a, b and c
i.e the value of a + b + c.
It's faster.
Your way after 3:14 is beautiful, I appreciate your idea.
Olympiad problem? Is this a joke?
Yes, it's nothing but indeed a joke
Probably its Olympiad for a 5 year school student. For them this is very hard
But if you solve it isolating every value, you'll get only the positive answer.
I started with, [ab = 100, bc = 200], [200 = 2 * 100], [bc = 2ab], cancel out the b's so that [c = 2a], then plug it into ca = 300 so that 2a^2 = 300 and solve for a.
ab=100 ; bc=200
So bc=2ab or
c=2a
ca=300=2a.a=2a*2
a*2=150
a= 5.(6)^2
b=100/(5×6^2)
b=(10×6^2)/3
c=10×6^2
a+b+c=(5+10/3+10)×6^2
a+b+c=(55×6^2)/3
A= 150/ кор.150
B= 100/ кор.150
С= 2× кор.150
Сумма 550/кор.150
bc/ab=200 /100 c=2a ac=a*2a 2a^2=300 a=5√6 c=10√6
ca/ab=300/100 c=3b b=(10/3)√6
I was imagining a longer possible approach, but I could not think of one.
a^2=150
b^2=200/3
a=5*sqrt(6)
b=10/3*sqrt(6)
c=10*sqrt(6)
Nossa 😅
bc/ab=c/a=2
ca*(c/a)=c^2=600
c=✓600=10✓6
a=300/10✓6=5✓6
b=100/5✓6=10✓6/3
😅😅😅
Or you multiply everything to get (abc)^2 then you deduce abc and a by dividing by bc and so on.
a=±5*sqrt(6); b=±(10/3)*sqrt(6); c=±10*sqrt(6).
I get
a= (3b)/2
b=√66.67
c=3b
I had to use a calculator to check the answer and it came surprisingly close.
ab=100.001
bc=200.001
ca=300.001
The longest way to solve IT)
1,100
2,50
4,25
5,20(100)
A, b, c are positive, so sum can't be negative.
Not necessarily.
ab, bc, and ca are positive, but
(-a)(-b)=ab
(-b)(-c)=bc
(-c)(-a)=ca
Beautiful.
A nice algebra problem
Исходные уравнения:
1. a *b = 100
2. b *c = 200
3. c *a = 300
Шаг 1: Найдем b через a
b = 100/a
Шаг 2: Подставим b в уравнение (2)
100/a *c = 200
100c = 200a
c = 2a
Шаг 3: Подставим c = 2a в уравнение (3)
c *a = 300
(2a) *a = 300
2a^2 = 300
a^2 = 150
a = √150= 5√6
Шаг 4: Найдем b через a
b = 100/a = 100/5√6= 20√6
Шаг 5: Найдем c через a
c = 2a = 2 *5√6= 10√6
Итак, решение:
a = 5√6
b = 20√6
c = 10√6
b -?
1)
a = 30/√6
b = 20/√6 (!!!)
c = 60/√6
2)
a = --30/√6
b = --20/√6 (!!!)
c = --60/√6
a+b+c = ± (30+20+60)/√6 = ±110/√6
Да... И кстати, все исходные могут быть либо положительными, либо отрицательными.
Произведения отрицательных также дают положительный результат.
Muy interesante.
A=√150
B=100÷√150
C =2√150
Solved in mind without pen papaer😂😂
Thanks
(bc)/(ab) = 2 => c = 2a
ac = 300 = 2a² => a = ± √150
=> c = ± 2√150
ab = 100 => b = ± 100/√150
a + b + c = ± (3√150 + 100/√150)
a + b + c = ± (550/√150)
a + b + c = ± [550/(5√6)]
a + b + c = ± (110/√6)
a + b + c = ± (110√6/6)
*a + b + c = ± (55√6/3)*
a + b + c = ± 110/√6
Зачем что-то ещё?
Здесь два действия, а потом уже аж три.
300
By this way, you will need one week to solve an exams, Faster sir,
Легко
ab = 100
bc = 200
ca = 300
===========
Metode I
ab*bc*ca = 100*200*300 = 100*6*10000
(abc)² = 100*6*10000
abc = ±10*√6*100 = ±1000√6
a = ab*ca/abc = 100*300/(±1000√6)
a = ±30/√6
b = ab*bc/abc = 100*200/(±1000√6)
b = ±20/√6
c = bc*ca/abc = 200*300/(±1000√6)
c = ±60/√6
a + b + c = ±(30+20+60)/√6
a + b + c = ±110/√6
==========
Methode 2
ab/bc = a/c = 100/200 = 1/2
a/c = 1/2 --> c = 2a
bc/ca = b/a = 200/300 = 2/3
b/a = 2/3 --> b = 2a/3
ca = 300
(2a)a = 300
2a² = 300
a² = 150
a = ±5√6
a + b + c = a + 2a/3 + 2a
a + b + c = 11a/3
a + b + c = 11(±5√6)/3
a + b + c = ±(55√6)/3
a + b + c = ±110/√6
=========
Method 3
As video
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Do we other method? Comment below 🙏🏻🙏🏻🙏🏻
Your methods give the right answer (method 2 seems simplest), but the convention seems to be to put the sqrts upstairs so (55/3)*sqrt(6) is preferable to 110/sqrt(6)
@@ZeroGravityDog thank you. . .
You're true. . .. 2nd methode is simplest than other one I know 🍵🍵🍵
See my answer for another way.
@@nenedillats2999you're true. . .
👌🏼👌🏼👌🏼
Why ab×ca/bc
Bec
Как-то это не убедительное для экзаменов ??? Простая трата времени находить ответ в таком виде !!! Атеперь встав и проверь-правильно-ли получил ???