Temple of Heaven

Algebra Level 4

The Xuanqiu altar in the Temple of Heaven in Beijing is a place for worshipping the heavens in ancient times, which consists of the upper, middle and lower layers, from inwards to outwards.

In the center of the upper layer is a circular slab (called the center stone). 9 9 fan-shaped slabs are built around the center stone to form the first ring. Each ring increases by 9 9 pieces of slabs outwards in each layer. The first ring of the next layer also has 9 9 pieces of slabs more than the last ring of the previous layer.

Knowing that the number of rings in each layer is the same, and the lower layer has 729 729 pieces of slabs more than the middle layer, how many slabs are there in the three layers (excluding the center stone)?

Source: Gaokao 2020, II


The answer is 3402.

Alice Smith by Alice Smith , 20, China

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2 solutions

David Vreken
Jul 28, 2020

The rings are in an arithmetic progression, so the first layer has a sum of L 1 = S n = 1 2 ( 9 n 2 + 9 n ) L_1 = S_{n} = \frac{1}{2}(9n^2 + 9n) slabs for n n rings in its layer.

The first and second layers have a sum of L 1 + L 2 = S 2 n = 1 2 ( 9 ( 2 n ) 2 + 9 ( 2 n ) ) L_1 + L_2 = S_{2n} = \frac{1}{2}(9(2n)^2 + 9(2n)) slabs for n n rings in each layer, so the second layer alone has L 2 = ( L 1 + L 2 ) − L 1 = S 2 n − S n = 1 2 ( 9 ( 2 n ) 2 + 9 ( 2 n ) ) − 1 2 ( 9 n 2 + 9 n ) L_2 = (L_1 + L_2) - L_1 = S_{2n} - S_{n} = \frac{1}{2}(9(2n)^2 + 9(2n)) - \frac{1}{2}(9n^2 + 9n) slabs.

The first, second, and third layers have a sum of L 1 + L 2 + L 3 = S 3 n = 1 2 ( 9 ( 3 n ) 2 + 9 ( 3 n ) ) L_1 + L_2 + L_3 = S_{3n} = \frac{1}{2}(9(3n)^2 + 9(3n)) slabs for n n rings in each layer, so the third layer alone has L 3 = ( L 1 + L 2 + L 3 ) − ( L 1 + L 2 ) = S 3 n − S 2 n = 1 2 ( 9 ( 3 n ) 2 + 9 ( 3 n ) ) − 1 2 ( 9 ( 2 n ) 2 + 9 ( 2 n ) ) L_3 = (L_1 + L_2 + L_3) - (L_1 + L_2) = S_{3n} - S_{2n} = \frac{1}{2}(9(3n)^2 + 9(3n)) - \frac{1}{2}(9(2n)^2 + 9(2n)) slabs.

The difference between the third and second layer is 729 729 , so L 3 − L 2 = ( 1 2 ( 9 ( 3 n ) 2 + 9 ( 3 n ) ) − 1 2 ( 9 ( 2 n ) 2 + 9 ( 2 n ) ) ) − ( 1 2 ( 9 ( 2 n ) 2 + 9 ( 2 n ) ) − 1 2 ( 9 n 2 + 9 n ) ) = 9 n 2 = 729 L_3 - L_2 = (\frac{1}{2}(9(3n)^2 + 9(3n)) - \frac{1}{2}(9(2n)^2 + 9(2n))) - (\frac{1}{2}(9(2n)^2 + 9(2n)) - \frac{1}{2}(9n^2 + 9n)) = 9n^2 = 729 , which solves to n = 9 n = 9 for n > 0 n > 0 .

There are therefore n = 9 n = 9 rings per layer, for a total of S 3 n = 1 2 ( 9 ( 3 n ) 2 + 9 ( 3 n ) ) = 1 2 ( 9 ( 3 ⋅ 9 ) 2 + 9 ( 3 ⋅ 9 ) ) = 3402 S_{3n} = \frac{1}{2}(9(3n)^2 + 9(3n)) = \frac{1}{2}(9(3 \cdot 9)^2 + 9(3 \cdot 9)) = \boxed{3402} slabs.

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Side note: It's interesting to know that since 9 is a noble number in China, which represents Yang, it makes sense to have 9 rings for each layer, as a check. Anyway, it's a good word problem in Gaokao, and it needs patience in such a short time :)

Alice Smith - 10 months, 2 weeks ago

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Yes, it's a nice problem!

David Vreken - 10 months, 2 weeks ago
Chew-Seong Cheong
Jul 29, 2020

Let the number of rings in a layer be n n . Then there is a total of 3 n 3n rings. Numbering the rings from 1 1 to 3 n 3n from the first ring of upper layer to the last ring of lower layer. We note that the numbers of slabs in a ring are in an arithmetic progression , where the k k th term or the number of slabs in k k th ring is a k = 9 k a_k = 9k .

Then the number of slabs in the middle layer, S 2 = ∑ k = n + 1 2 n a k = n ( a n + 1 + a 2 n ) 2 = 9 n ( 3 n + 1 ) 2 \displaystyle S_2 = \sum_{k=n+1}^{2n} a_k = \frac {n(a_{n+1}+a_{2n})}2 = \frac {9n(3n+1)}2 .

Similarly, the number of slabs in the lower layer, S 3 = ∑ k = 2 n + 1 3 n a k = n ( a 2 n + 1 + a 3 n ) 2 = 9 n ( 5 n + 1 ) 2 \displaystyle S_3 = \sum_{k=2n+1}^{3n} a_k = \frac {n(a_{2n+1}+a_{3n})}2 = \frac {9n(5n+1)}2 .

Given that S 3 − S 2 = 729 S_3 - S_2 = 729 , ⟹ 9 n ( 5 n + 1 ) 2 − 9 n ( 3 n + 1 ) 2 = 9 n 2 = 729 ⟹ n = 9 \implies \dfrac {9n(5n+1)}2 - \dfrac {9n(3n+1)}2 = 9n^2 = 729 \implies n = 9 ( Should of thought of it. ).

Then the total number of slabs in the three layers S = ∑ k = 1 27 a k = 27 × 9 ( 1 + 27 ) 2 = 3402 \displaystyle S = \sum_{k=1}^{27} a_k = \frac {27\times 9(1+27)}2 = \boxed{3402} .

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