Apollonius' Riddle
Apollonius, the famous mathematician, was invited by the King of Perga for a geometry session. Apollonius first presented the painting of the 3 squares rearranged to form a triangle with a median splitting the largest side, as shown above.
Apollonius : O my great King, behold this beautiful diagram! The Lord of the North has 2 pieces of red square lands while the Lord of the South has 1 larger blue square land. Their sides including the median all have lengths in whole integers.
King : Beautiful indeed, my teacher!
Apollonius : Then does thy greatness know who has more land?
King : I've heard that the Lord of the South has more (blue) land by 26.
Apollonius : Now then all side lengths are known. Please tell me, my wise one. What is the perimeter of this triangle?
The answer is 32.
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2 solutions
Why have you typed 1 5 0 up there; shouldnt it be 2 6 ?
Now, consider these triplets for
(
a
,
b
,
c
)
(
3
,
1
7
,
1
8
)
;
(
5
,
2
5
,
2
6
)
;
(
7
,
3
7
,
3
8
)
;
(
9
,
5
3
,
5
4
)
;
(
1
1
,
7
3
,
7
4
)
and so on; why are all these invalid?
Now, listen I think where you (or me) have messed up is;
the lengths of the median and
m
need not be whole numbers; which you have assumed to get to the answer
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In the question, Apollonius said it had to be integer.
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But; he had said about the sides of the square (and consequently the triangle); and not the median of the triangle.
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@Yatin Khanna – Sorry. Just edited.
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@Worranat Pakornrat – Seems fine now.
L e t a , b , c b e t h e s i d e s , c > a , b , a n d d t h e m e d i a n a l l i n t e g e r s . ∴ a 2 + b 2 = c 2 − 2 6 . . . . . . . ( ∗ ∗ ) B u t d 2 = 2 1 ∗ ( a 2 + b 2 ) − 4 1 ∗ c 2 = 2 1 ∗ ( c 2 − 2 6 ) − 4 1 ∗ c 2 = 4 1 ∗ c 2 − 1 3 . ⟹ 4 1 ∗ c 2 − d 2 = 1 3 . ⟹ ( 2 1 c − d ) ( 2 1 c + d ) = 1 3 . S i n c e c , a n d d a r e i n t e g e r s , o n l y s o l u t i o n i s ( 2 1 c − d ) = 1 , a n d ( 2 1 c + d ) = 1 3 . ∴ c = 1 4 . S o f r o m ( ∗ ∗ ) , a 2 + b 2 = 1 9 6 − 2 6 = 1 7 0 . S i n c e t h i s i s e v e n , b o t h a a n d b a r e e v e n o r o d d . H o w e v e r t h e s u m o f t w o s q a u r e s h a s z e r o i n u n i t d i g i t . S i n c e 2 1 7 0 = 8 5 N O T a s q u a r e . W L O G a < b . ∴ ( A ) { 5 ∣ a a n d 5 ∣ b } , o r ( B ) { ( 4 o r 6 ) ∣ a a n d ( 6 o r 4 ) ∣ b } , o r ( C ) { ( 1 o r 9 ) ∣ a a n d ( 9 o r 1 ) ∣ b } ( A ) 1 7 0 − 5 2 = 1 4 5 N O T a s q u a r e . N o o t h e r c o m b i n a t i o n h e r e c a n g i v e a t o t a l o f 1 7 0 . ( B ) 1 7 0 − 4 2 = 1 5 4 , N O T a s q u a r e , 1 7 0 − 6 2 = 1 3 4 N O T a s q u a r e . N o o t h e r c o m b i n a t i o n h e r e c a n g i v e a t o t a l o f 1 7 0 . ( C ) 1 7 0 − 1 2 = 1 6 9 = 1 3 2 . s a t i s f y ( ∗ ∗ ) , B u t 1 , 1 3 , 1 4 i s a s t . l i n e . 1 7 0 − 3 2 = 1 6 1 N O T a s q u a r e . . . . . 1 7 0 − 7 2 = 1 2 1 = 1 1 2 . s a t i s f y ( ∗ ∗ ) , S o a = 7 a n d b = 1 1 . A n s w e r . 1 4 + 1 1 + 7 = 3 2 .
You can't draw a triangle with sides 1, 13, 14: it will be a straight line.
Nice casework here.
More generally, we can use gaussian integers to help solve a 2 + b 2 = n .
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Thank you for pointing out, a new for me.
According to Apollonius' theorem, for a triangle, with side lengths a , b , c , where c = 2 m , and the median length d as shown, a 2 + b 2 = 2 ( d 2 + m 2 ) .
This can be rewritten as a 2 + b 2 − c 2 = 2 ( d 2 − m 2 ) = 2 ( d − m ) ( d + m ) .
Or c 2 − a 2 − b 2 = 2 ( m − d ) ( m + d ) .
In the question, the blue land has larger area than the red ones by 2 6 .
Hence, c 2 − a 2 − b 2 = 2 6 = 2 ( m − d ) ( m + d ) .
1 3 = ( m − d ) ( m + d ) .
Since all side lengths are all in integers, there are only one possibility: 1 3 = 1 ⋅ 1 3 .
Then if we add the factors together, ( m − d ) + ( m + d ) = 2 m = c = 1 + 1 3 = 1 4 .
Then m = 7 and d = 6 .
Thus, a 2 + b 2 = 2 ( d 2 + m 2 ) = 2 ( 6 2 + 7 2 ) = 1 7 0 .
Then there are two possibilities: 1 7 0 = 1 2 + 1 3 2 = 7 2 + 1 1 2 .
However, if a = 1 ; b = 1 3 ; c = 1 4 , only a straight line can be drawn.
Therefore, a = 7 ; b = 1 1 ; c = 1 4 , and the perimeter is 7 + 1 1 + 1 4 = 3 2 .