Tricking the Magician

Let $(a, b, c, L, W)$ be the sides of the $3$ squares in increasing order, the length, and the width of the rectangle. Then after the magician has explained the first set of conditions, only the following are possible

$(1,2,4,7,3)$
$(1,3, 9,13,7)$
$(1,4,9,14,7)$
$(4,6,9,19,7)$

Then after the magician explained that one cannot determine the area of the rectangle even after knowing with the width is, then the first possibility is eliminated, and the remaining following are possible

$(1,3, 9,13,7)$
$(1,4,9,14,7)$
$(4,6,9,19,7)$

Then after the magician explained that if one knew the side length of any of the three squares, one still cannot determine the area of the rectangle, that leaves the only possibility

$(1,4,9,14,7)$

since it's the only one of the three where the side of each of the three squares matches the side of a square in at least one other possibility. Hence, the area is $14 \cdot 7=98$

This is a shortcut summary of Worrant's solution as given near the end of it.

Relevant wiki: K-level thinking

Suppose $a, b, c$ be the side lengths of the three squares and $d$ be the width of the rectangle. With the same area, we can set up an equation as shown below:

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

From the given equation, it is clear that $a+b+c| a^2+b^2+c^2$ . Hence, $a^2+b^2+c^2 \equiv 0 \pmod{a+b+c}$ .

And $(a+b+c)^2 \equiv (a^2+b^2+c^2)+2(ab+bc+ca) \equiv 0 \pmod{a+b+c}$ .

If $a^2+b^2+c^2 \equiv 0 \pmod{a+b+c}$ , then $2(ab+bc+ca) \equiv 0 \pmod{a+b+c}$ .

Then $2((a+b)(b+c) - b^2) \equiv 2((-c)(-a)-b^2) \equiv 2(ac - b^2) \equiv 0 \pmod{a+b+c}$ .

Therefore, if $b^2 = ac$ , the equivalence will be true, and from all the digit squares, here are the possibilities:

$2^2 = 1\cdot 4$ . Thus, $\dfrac{1^2 + 2^2 + 4^2}{1+2+4} = \dfrac{21}{7} = 3$ .

$3^2 = 1\cdot 9$ . Thus, $\dfrac{1^2 + 3^2 + 9^2}{1+3+9} = \dfrac{91}{13} = 7$ .

$4^2 = 2\cdot 8$ . Thus, $\dfrac{2^2 + 4^2 + 8^2}{2+4+8} = \dfrac{84}{14} = 6$ .

$6^2 = 4\cdot 9$ . Thus, $\dfrac{4^2 + 6^2 + 9^2}{4+6+9} = \dfrac{133}{19} = 7$ .

Now we have explored the most obvious solutions, but are there more?

$$ Since $b \equiv -a-c \pmod{a+b+c}$ , then $b^2 \equiv (a+c)^2 \pmod{a+b+c}$ .

Hence, $2(ac - b^2) \equiv 2(ac - (a+c)^2) \equiv 2(a^2 + ac + c^2) \equiv 0 \pmod{a+b+c}$ .

Then $2(c-a)(a^2 + ac + c^2) \equiv 2(c^3 - a^3) \equiv 0 \pmod{a+b+c}$ .

Similarly, with the terms alternated, we will have $2(c^3 - b^3) \equiv 2(b^3 - a^3) \equiv 0 \pmod{a+b+c}$ .

For generality, let us have $a<b<c$ . This way, we will attempt to find the middle digit for all possible square summation by checking the upper or the lower bound from that number.

Then the case of $b=2$ is already checked as shown in the above solution.

$$ For $b=3$ , we will have $a=2$ left, and $2(3^3 - 2^3) \equiv 2\cdot 19 \equiv 0 \pmod{2+3+c}$ . Then $c$ must be $14$ to make the equivalence true, but it's not the desired digit, according to the problem.

And we have the triple $a=1$ , $b=3$ , and $c=9$ as shown in the above solution.

$$ Then for $b=4$ , if $a=1$ , $2(4^3 - 1^3) \equiv 126 \equiv 2\cdot 3\cdot 3\cdot 7 \equiv 0 \pmod{1+4+c}$ . Then $c$ can be $4$ or $9$ , but if we want all distinct digits, only $9$ applies: $\dfrac{1^2 + 4^2 + 9^2}{1+4+9} = \dfrac{98}{14} = 7$ .

And we have the triple $a=2$ , $b=4$ , and $c=8$ as shown in the above solution.

For $b=4$ , if $a=3$ , $2(4^3 - 3^3) \equiv 2\cdot 37 \equiv 0 \pmod{1+4+c}$ . Then $c$ must be $32$ , which is not applicable.

$$ For $b=5$ , if $a=1$ , $2(5^3 - 1^3) \equiv 2\cdot 124 \equiv 8\cdot 31 \equiv 0 \pmod{1+5+c}$ . Only $2$ or $25$ works for $c$ here, but $2$ only works for $c=4$ as shown above, and $25$ is not applicable.

For $b=5$ , if $a=2$ , $2(5^3 - 2^3) \equiv 2\cdot 117 \equiv 2\cdot 3\cdot 3\cdot 13 \equiv 0 \pmod{2+5+c}$ . Only $2$ or $6$ works for $c$ , but if we want the distinct digit, only $6$ applies: $\dfrac{2^2 + 5^2 + 6^2}{2+5+6} = \dfrac{65}{13} = 5$ .

Unfortunately, it's not a distinct quotient digit, so not a new solution for us.

For $b=5$ , if $a=3$ , $2(5^3 - 3^3) \equiv 2\cdot 98 \equiv 2\cdot 2\cdot 7\cdot 7 \equiv 0 \pmod{3+5+c}$ . Only $6$ applies for $c$ : $\dfrac{3^2 + 5^2 + 6^2}{3+5+6} = \dfrac{70}{14} = 5$ . Again, it's not a distinct quotient digit, so not a new solution for us.

For $b=5$ , if $a=4$ , $2(5^3 - 4^3) \equiv 2\cdot 61 \equiv 0 \pmod{4+5+c}$ . No digit $c$ exists for this case.

$$ For $b=6$ , if $c=7$ , $2(7^3 - 6^3) \equiv 2\cdot 127 \equiv 0 \pmod{a+6+7}$ . No digit $a$ exists for this case.

For $b=6$ , if $c=8$ , $2(8^3 - 6^3) \equiv 2\cdot 296 \equiv 2\cdot 2\cdot 2\cdot 74 \equiv 0 \pmod{a+6+8}$ . No digit $a$ exists for this case.

And we have the triple $a=4$ , $b=6$ , and $c=9$ as shown in the above solution.

$$ For $b=7$ , if $c=8$ , $2(8^3 - 7^3) \equiv 2\cdot 169 \equiv 2\cdot 13\cdot 13 \equiv 0 \pmod{a+7+8}$ . No digit $a$ exists for this case.

For $b=7$ , if $c=9$ , $2(9^3 - 7^3) \equiv 2\cdot 386 \equiv 2\cdot 2\cdot 193 \equiv 0 \pmod{a+7+9}$ . No digit $a$ exists for this case.

$$ For $b=8$ , if $c=9$ , $2(9^3 - 8^3) \equiv 2\cdot 217 \equiv 2\cdot 7\cdot 31\equiv 0 \pmod{a+8+9}$ . No digit $a$ exists for this case.

$$

Thus, we can conclude that the only solution $(a,b,c)$ are: $(1,2,4)$ ; $(1,3,9)$ ; $(1,4,9)$ ; $(2,4,8)$ ; and $(4,6,9)$ .

And the only applicable quotient digits or the widths would include: $3$ , $6$ , & $7$ .

From the magician's statement, even if we know the width, we still can't be sure of the area of the rectangle. Therefore, the width can't be $3$ or $6$ as each has only one side length correlation. Thus, we can deduce that the width is $7$ .

Then for the width of $7$ , we will have these possible $(a,b,c)$ combinations: $(1,3,9)$ ; $(1,4,9)$ ; $(4,6,9)$ .

Again from the magician's last information, even if we know just any of the square's side, we won't be able to work out the whole area. That means such square's side would need to be repeated in these combinations, and it is obvious that the digits $1$ , $4$ , & $9$ appear in more than one possible outcome. In other words, if we are told that the square side is $3$ or $6$ , we can work out the rest of the sides, which is not the case.

As a result, the area of the rectangle = $7(1+4+9) = 1^2 + 4^2 + 9^2 = \boxed{98}$ .

I have a Question.

If I set up the equation as

a^2+b^2+c^2 = (a+b+c)x

I get x = 7 and possible solutions are (1,3,9), (1,4,9), (3,6,9), (4,6,9).

In any of these cases, revealing x = 7 does not result in finding the answer since it could be any of these four. Revealing any one of these squares' side length does not result in finding the answer since for any of the side lengths (1,3,4,6,9) has at least two cases it can be true. How can you deduce further from this point that it is the (1,4,9)? Thank you.