MHB Can you Solve This Putnam 2012 Problem about Acute Triangles?

  • Thread starter Thread starter jakncoke1
  • Start date Start date
jakncoke1
Messages
48
Reaction score
1
Lets take a crack at em.

Prob A1: Let $d_1,...,d_{12}$ be 12 real numbers in the interval (1,12), show that there exists indicides $i,j,k$, such that $d_i,d_j,d_k$ are side lengths of an acute triangle.
 
Mathematics news on Phys.org
This is actually quite easy.
WLOG, Let $d_1 \leq ... \leq d_{12}$ . If we note that for sides a,b,c , the triangle being acute is equivalent to $a^2+b^2 > c^2$ $a \leq b \leq c$

Assume that there exists no $ i < k < j$ such that $(d_j)^2< (d_i)^2 + (d_k)^2.$ There fore, $d_1^2 + d_2^2 \leq d_3^2$ , since $d_k > 1 for all k. d_3^2 > 2$.
Then $d_4^2 > d_3^2 + d_2^2 = d^4 > 2 + 1$
$d_5^2 > d_4^2 + d_3^2 = 3 + 2$
so $d_p ^2 > F(p)$ where F(p) is the pth fibonacci number.
so $d_{12}^2 > F(12)$
or $d_{12}^2 > 144, d_{12} > 12$, which is a contradiction.
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

3
Replies
100
Views
11K
Replies
3
Views
2K
Replies
125
Views
19K
Replies
48
Views
11K
Replies
39
Views
13K
Replies
1
Views
3K
Back
Top