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# Math Revolution DS Expert - Ask Me Anything about GMAT DS

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10135
Own Kudos [?]: 16999 [1]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10135
Own Kudos [?]: 16999 [1]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10135
Own Kudos [?]: 16999 [1]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10135
Own Kudos [?]: 16999 [1]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
1
Kudos
[GMAT math practice question]

(Algebra) What is the value of $$a^2 + b^2 + c^2 – ab – bc - ca$$?

1) $$a-b=-1$$

2) $$b-c=\sqrt{2}$$
Math Revolution GMAT Instructor
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MathRevolution wrote:
[GMAT math practice question]

(Algebra) $$p = \frac{a}{(a+b)(a+c)}+\frac{b}{(b+c)(b+a)}+\frac{c}{(c+a)(c+b)}, q = \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}.$$ What is $$p-q$$?

1) $$a, b,$$ and $$c$$ are integers.

2) $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 0.$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

$$p = \frac{a}{(a+b)(a+c)}+\frac{b}{(b+c)(b+a)}+\frac{c}{(c+a)(c+b)}$$

$$= \frac{a(b+c)}{(a+b)(a+c)(b+c)}+\frac{b(c+a)}{(b+c)(b+a)(c+a)}+\frac{c(a+b)}{(c+a)(c+b)(a+b)}$$(getting a common denominator)

$$= \frac{ab+ca+bc+ab+ca+bc}{(a+b)(b+c)(c+a)}$$ (multiplying through the brackets and combining in one fraction)

$$= \frac{2(ab+ca+bc)}{(a+b)(b+c)(c+a)}$$ (adding like terms and taking out a common factor of $$2$$)

$$q = \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}$$

$$= \frac{b+c+a-a}{a}+\frac{c+a+b-b}{b}+\frac{a+b+c-c}{c}$$ (adding and subtracting variables in order to get a numerator of $$a + b + c$$)

$$= \frac{a+b+c}{a}-\frac{a}{a}+\frac{a+b+c}{b}-\frac{b}{b}+\frac{a+b+c}{c}-\frac{c}{c}$$ (separating fractions)

$$= \frac{a+b+c}{a}-1+\frac{a+b+c}{b}-1+\frac{a+b+c}{c}-1$$ (simplifying fractions)

$$= (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3$$ (adding the constants and taking out a common factor of $$a + b + c$$)

Condition 2)
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$$

$$=> \frac{bc}{abc}+\frac{ca}{abc}+\frac{ab}{abc}=0$$ (getting a common denominator)

$$=> \frac{ab+bc+ca}{abc}=0$$ (combining into one fraction)

$$=> ab + bc + ca = 0$$ (multiplying both sides by $$abc$$)

Then, we have $$p = \frac{2(ab+ca+bc)}{(a+b)(b+c)(c+a)} = \frac{0}{(a+b)(b+c)(c+a)} = 0$$ and $$q = (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3 = (a+b+c)*0-3 = -3.$$

Thus, $$p – q = 0 – (-3) = 3.$$

Since condition 2) yields a unique solution, it is sufficient.

Condition 1)

Since condition 1) does not yield a unique solution, it is not sufficient.

Originally posted by MathRevolution on 25 Mar 2020, 06:56.
Last edited by MathRevolution on 12 Jan 2022, 03:09, edited 1 time in total.
Math Revolution GMAT Instructor
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[GMAT math practice question]

(Function) $$f(x)$$ and $$g(x)$$ are functions. What is the value $$f(3) + g(3)$$?

1) $$f(x + g(y)) = ax + y + 1$$ where a is a constant.

2) $$f(0) = -2, g(0) = 1.$$
Math Revolution GMAT Instructor
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Own Kudos [?]: 16999 [1]
Given Kudos: 4
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1
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MathRevolution wrote:
[GMAT math practice question]

(Algebra) What is the value of $$a^2 + b^2 + c^2 – ab – bc - ca$$?

1) $$a-b=-1$$

2) $$b-c=\sqrt{2}$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since we have $$3$$ variables ($$x, y,$$ and $$z$$) and $$0$$ equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

The question $$a^2 + b^2 + c^2 – ab – bc - ca$$ is equivalent to $$\frac{1}{2}{(a-b)^2+(b-c)^2+(c-a)^2}$$ for the following reason
$$a^2 + b^2 + c^2 – ab – bc - ca$$
$$= \frac{1}{2}{2a^2 + 2b^2 + 2c^2 – 2ab – 2bc - 2ca}$$
$$= \frac{1}{2}{(a^2-2ab + b^2)+(b^2-2bc + c^2)+(c^2 - 2ca+a^2)}$$
$$=\frac{1}{2}{(a-b)^2+(b-c)^2+(c-a)^2}$$

Conditions 1) & 2)
Since we have $$a-b=-1$$ and $$b-c=\sqrt{2}$$, we have $$c - a = -( c – b + b – a ) = -( √2 + (-1)) = 1 - √2.$$

$$a^2 + b^2 + c^2 – ab – bc - ca$$

$$= \frac{1}{2}{(a-b)^2+(b-c)^2+(c-a)^2}$$

$$= \frac{1}{2}{(-1)^2+\sqrt{2}^2+(1-\sqrt{2})^2}$$

$$= \frac{1}{2}(1+2+3-2\sqrt{2})$$

$$= 3 - \sqrt{2}$$

Since both conditions together yield a unique solution, they are sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.

Originally posted by MathRevolution on 26 Mar 2020, 05:09.
Last edited by MathRevolution on 12 Jan 2022, 03:09, edited 1 time in total.
Math Revolution GMAT Instructor
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[GMAT math practice question]

(Geometry) We have a square $$□ABCD$$, and $$△AEF$$ is a triangle inscribed in $$□ABCD$$. What is the length of $$BE$$?

1) $$□ABCD$$ is a square with $$AB = 10$$

2) $$△AEF$$ is an equilateral triangle.

Attachment:

3.26(ds).png [ 5.63 KiB | Viewed 1533 times ]
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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Own Kudos [?]: 16999 [1]
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1
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MathRevolution wrote:
[GMAT math practice question]

(Function) $$f(x)$$ and $$g(x)$$ are functions. What is the value $$f(3) + g(3)$$?

1) $$f(x + g(y)) = ax + y + 1$$ where a is a constant.

2) $$f(0) = -2, g(0) = 1.$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since we have $$3$$ variables ($$x, y,$$ and $$a$$) and $$0$$ equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
When we substitute $$0$$ for $$y$$, we have$$f(x + g(0)) = f(x+1) = ax + 1.$$

Then, we have $$f(0) = f((-1) + 1) = a(-1) + 1 = 1 – a = -2$$ or $$a = 3.$$

$$f(3) = f(2+1) = 3*2+1 = 7.$$

We have $$f(x + g(y) – 1 + 1) = 3(x + g(y) - 1) + 1 = 3x + 3g(y) - 2 = 3x + y + 1$$ or $$3g(y) = y – 3$$.

Then $$g(y) = \frac{y}{3} + 1$$ and $$g(3) = 2.$$

Thus $$f(3) + g(3) = 7 + 2 = 9.$$

Since both conditions together yield a unique solution, they are sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
Math Revolution GMAT Instructor
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[GMAT math practice question]

(Geometry) The figure shows the right triangle $$ABC$$ and inscribed circle, which is tangential at $$D$$. $$AD = p$$ and $$CD = q.$$ What is the area of $$△ABC$$?

1) $$p = 2, q = 3.$$

2) The radius of the inscribed circle is $$1$$.

Attachment:

3.27DS.png [ 9.45 KiB | Viewed 1503 times ]
Math Revolution GMAT Instructor
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MathRevolution wrote:
[GMAT math practice question]

(Geometry) We have a square $$□ABCD$$, and $$△AEF$$ is a triangle inscribed in $$□ABCD$$. What is the length of $$BE$$?

1) $$□ABCD$$ is a square with $$AB = 10$$

2) $$△AEF$$ is an equilateral triangle.

Attachment:
The attachment 3.26(ds).png is no longer available

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have a triangle and a square, we have 3 variables and 1 variable for the triangle and the square, respectively. Since we have 4 variables and 0 equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

Attachment:

3.26ds(a).png [ 9.37 KiB | Viewed 1496 times ]

Assume $$BE = x (0 < x < 10).$$

Then we have $$BE = DF$$ since the triangle $$AEF$$ is equilateral.

$$AE^2 = 100 + x^2 = EF^2 = (10 - x)^2 + (10 - x)^2.$$

Then we have
$$100 + x^2 = (10 - x)^2 + (10 - x)^2$$

$$100 + x^2 = 2(10 – x)^2$$ (adding like terms)

$$100 + x^2 = 2(10 – x)(10 – x)$$

$$100 + x^2 = 2(100 – 10x – 10x + x^2)$$ (foiling out the brackets)

$$100 + x^2 = 2x^2 – 40x + 200$$ (multiplying $$2$$ through the bracket and combining like terms)

$$x^2 – 40x + 100 = 0$$ (bringing all terms to one side)

Thus, we have $$x = 20 ± 10√3$$. (using the quadratic formula)

But we have $$x = 20 - 10√3$$ since $$0 < x < 10.$$

Since both conditions together yield a unique solution, they are sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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1
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MathRevolution wrote:
[GMAT math practice question]

(Geometry) The figure shows the right triangle $$ABC$$ and inscribed circle, which is tangential at $$D$$. $$AD = p$$ and $$CD = q.$$ What is the area of $$△ABC$$?

1) $$p = 2, q = 3.$$

2) The radius of the inscribed circle is $$1$$.

Attachment:
The attachment 3.27DS.png is no longer available

=>

Attachment:

3.27ds(a).png [ 17.76 KiB | Viewed 1495 times ]

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

We can put $$BC = BF = x.$$

Put $$BC = a = q + x, AC = b = p + q$$ and $$AB = c = p + x.$$

Then the area of triangle $$ABC = (\frac{1}{2})ac$$ and $$x = \frac{(a-b+c)}{2}, p = \frac{(b-(a-c))}{2}$$ and $$q = \frac{(b+(a-c))}{2}$$

Then we have
$$pq = {\frac{(b-(a-c))}{2}}{\frac{(b+(a-c))}{2}}$$

$$pq = \frac{{b^2 + b(a-c) – b(a-c) - (a-c)^2}}{4}$$

$$pq = \frac{(b^2 – (a-c)^2)}{4 }$$

$$pq = \frac{{b^2 – (a-c)(a-c)}}{4}$$

$$pq = \frac{{b^2 – (a^2 – ac – ac + c^2)}}{4}$$

$$pq = \frac{(b^2 - a^2 - c^2 + 2ac)}{4 }$$

$$pq = \frac{ac}{2}.$$

Thus, the area of the triangle $$ABC$$ is $$pq$$, and condition 1) is sufficient.

Condition 2)

Since there are a lot of possibilities for $$p$$ and $$q$$, condition 2) does not yield a unique solution, and it is not sufficient.

Math Revolution GMAT Instructor
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[GMAT math practice question]

(Algebra) $$x, y$$ and $$z$$ are real numbers with $$xyz = 1$$. What is the value of $$(x - 1)(y - 1)(z - 1)$$?

1) $$x + y + z = 3$$

2) $$xy + yz + zx = -4$$
Math Revolution GMAT Instructor
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[GMAT math practice question]

(Number Property) $$N$$ is an integer. Is $$N$$ a perfect square?

1) $$N$$ is $$1$$ greater than the product of $$4$$ consecutive integers.

2) $$N$$ is a summation of squares of $$4$$ consecutive odd integers.
Math Revolution GMAT Instructor
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Own Kudos [?]: 16999 [1]
Given Kudos: 4
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MathRevolution wrote:
[GMAT math practice question]

(Algebra) $$x, y$$ and $$z$$ are real numbers with $$xyz = 1$$. What is the value of $$(x - 1)(y - 1)(z - 1)$$?

1) $$x + y + z = 3$$

2) $$xy + yz + zx = -4$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

$$(x - 1)(y - 1)(z - 1)$$

$$= (x – 1)(yz – y – z + 1)$$

$$= xyz – xy – xz + x – yz + y + z - 1$$

$$= xyz – (xy + yz + zx) + (x + y + z) – 1$$

Since we have $$x + y + z = 3$$ and $$xy + yz + zx = -4$$ from both conditions 1) and 2), we have $$(x - 1)(y - 1)(z - 1) = xyz – (xy + yz + zx) + (x + y + z) – 1 = 1 – (-4) + 3 – 1 = 7$$.

Since both conditions together yield a unique solution, they are sufficient.

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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[GMAT math practice question]

(Number Property) $$a, b,$$ and $$c$$ are integers. Is $$2(a^4 + b^4 + c^4)$$ a perfect square?

1) $$a = 1, b = 1$$, and $$c = -2$$

2) $$a + b + c = 0$$
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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Own Kudos [?]: 16999 [1]
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MathRevolution wrote:
[GMAT math practice question]

(Number Property) $$N$$ is an integer. Is $$N$$ a perfect square?

1) $$N$$ is $$1$$ greater than the product of $$4$$ consecutive integers.

2) $$N$$ is a summation of squares of $$4$$ consecutive odd integers.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have $$1$$ variable ($$N$$) and $$0$$ equations, D is most likely the answer. So, we should consider each condition on its own first.

Condition 1)

Assume $$N$$ is $$1$$ greater than a product of four consecutive integers, $$x, x+1, x+2,$$ and $$x+3$$ where $$x$$ is an integer.
We have
$$N = x(x + 1)(x + 2)(x + 3) + 1$$

$$N = x(x + 3)(x + 1)(x + 2) + 1$$

$$N = (x^2 + 3x)(x^2 + 3x + 2) + 1$$

$$N = (x^2 + 3x)^2 + 2(x^2 + 3x) + 1$$

$$N = (x^2 + 3x + 1)^2$$

Thus, $$N$$ is a perfect square.

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)

If $$N = 1 + 3 + 5 + 7 = 16$$, then $$N$$ is a perfect square and the answer is ‘yes’.

If $$N = 3 + 5 + 7 + 9 = 24$$, then $$N$$ is not a perfect square and the answer is ‘no’.

Since condition 2) does not yield a unique solution, it is not sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations,” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A, B, C, or E.

Originally posted by MathRevolution on 02 Apr 2020, 01:19.
Last edited by MathRevolution on 01 Apr 2021, 02:53, edited 1 time in total.
Math Revolution GMAT Instructor
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[GMAT math practice question]

(Function) What is the value of $$f(2019)$$?

1) $$f(3) = 5$$

2) $$f(x+2) = \frac{f(x) - 1}{f(x) + 1}$$
Math Revolution GMAT Instructor
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MathRevolution wrote:
[GMAT math practice question]

(Number Property) $$a, b,$$ and $$c$$ are integers. Is $$2(a^4 + b^4 + c^4)$$ a perfect square?

1) $$a = 1, b = 1$$, and $$c = -2$$

2) $$a + b + c = 0$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Condition 1)

Since we have $$a = 1, b = 1,$$ and $$c = 2$$, we have

$$2(a^4 + b^4 + c^4) = 2(1^4 + 1^4 + 2^4)$$

$$= 2(1 + 1 + 16)$$

$$= 2*18$$

$$= 36.$$

$$2(a^4 + b^4 + c^4) = 36$$ is a perfect square and the answer is ‘yes’.

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)

Since $$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca),$$

we have $$a^2 + b^2 + c^2 = -2(ab + bc + ca).$$

Then, rearranging the second formula gives us:
$$(a^2 + b^2 + c^2)^2 = (-2(ab + bc + ca))^2$$

$$= 4(ab + bc + ca)(ab + bc + ca)$$

$$= 4(a^2b^2 + ab^2c + a^2bc + ab^2c + b^2c^2 + abc^2 + a^2bc + abc^2 + a^2c^2)$$

$$= 4((a^2b^2 + b^2c^2 + a^2c^2 + 2ab^2c + 2a^2bc + 2abc^2)$$

$$= 4(a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c))$$

$$= 4(a^2b^2 + b^2c^2 + c^2a^2),$$since $$a + b + c = 0$$

Following the pattern in the first equation gives us:
$$(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2)$$

We now have two equations:
$$(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 +b^2c^2+ c^2a^2)$$

$$(a^2 + b^2 + c^2)^2 = 4(a^2b^2 + b^2c^2 + c^2a^2)$$

Combining the two equations gives us:
$$a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2+ c^2a^2) = 4(a^2b^2 + b^2c^2 + c^2a^2)$$

$$a^4 + b^4 + c^4 = 2(a^2b^2 + b^2c^2 + c^2a^2)$$

$$2(a^4 + b^4 + c^4) = 4(a^2b^2 + b^2c^2 + c^2a^2) = (a^2 + b^2 + c^2)^2.$$

Thus, $$2(a^4 + b^4 + c^4)$$ is a perfect square.

Since condition 2) yields a unique solution, it is sufficient.

This question is a CMT 4(B) question: condition 1) is easy to work with, and condition 2) is difficult to work with. For CMT 4(B) questions, D is most likely the answer.

Originally posted by MathRevolution on 03 Apr 2020, 02:44.
Last edited by MathRevolution on 01 Apr 2021, 02:52, edited 1 time in total.
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