Is This Another Triangle Inequality?

Geometry Level 4

Find the smallest k k such that the sum of the legs of a right triangle never exceeds k \sqrt{k} times the hypotenuse of the triangle.

Give your answer to 3 decimal places.


The answer is 2.

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5 solutions

Hung Woei Neoh
Jun 3, 2016

Relevant wiki: Solving Triangles - Problem Solving - Medium

Let the legs of a triangle be a a , b b and the hypotenuse be c c

We are looking for the smallest k k such that

a + b ≤ k c ( a + b ) 2 ≤ ( k c ) 2 a 2 + 2 a b + b 2 ≤ k c 2 a 2 + 2 a b + b 2 ≤ k ( a 2 + b 2 ) k ( a 2 + b 2 ) − a 2 − b 2 ≥ 2 a b ( k − 1 ) ( a 2 + b 2 ) ≥ 2 a b a+b \leq \sqrt{k}c\\ (a+b)^2 \leq (\sqrt{k}c)^2\\ a^2+2ab+b^2 \leq kc^2\\ a^2+2ab+b^2 \leq k(a^2+b^2)\\ k(a^2+b^2) -a^2-b^2 \geq 2ab\\ (k-1)(a^2+b^2) \geq 2ab

Now, since a 2 > 0 , b 2 > 0 a^2 > 0,\;b^2 > 0 , from AM-GM:

a 2 + b 2 2 ≥ a 2 b 2 a 2 + b 2 ≥ 2 a b \dfrac{a^2+b^2}{2} \geq \sqrt{a^2b^2}\\ a^2+b^2 \geq 2ab

Therefore,

( k − 1 ) ( a 2 + b 2 ) − ( a 2 + b 2 ) ≥ 2 a b − 2 a b ( k − 2 ) ( a 2 + b 2 ) ≥ 0 k − 2 ≥ 0 k ≥ 2 (k-1)(a^2+b^2) -(a^2+b^2) \geq 2ab - 2ab\\ (k-2)(a^2+b^2) \geq 0\\ k-2 \geq 0\\ k \geq 2

The smallest natural number of k = 2 k=\boxed{2}

15 Helpful 0 Interesting 0 Brilliant 1 Confused

that was cool

abhishek alva - 5 years ago

I think there is a mistake.if a>b and c> b then a-c is not necessarily greater than 0

firoz khan - 5 years ago

(k-1)(a/b +b/a)≥2 is true for k-1≥1

firoz khan - 5 years ago

I have it similar, but after k-1>=2ab/(a^2 +b^2), I replaced it with 2ch/c^2, with h as the height. And since h<=c/2 (Thales) it gives you k-1>= 2 c 1/2*c / (c^2) =1 => k>=2

Bibin Muttappillil - 5 years ago

You subtracted the second equation from the first. If we subtract the first from the second, we get k ≤ 2 k \leq 2 which is reverse of what you tried to prove.

Atomsky Jahid - 1 year, 1 month ago
Atomsky Jahid
Jun 14, 2016

The square of a real number is always greater than zero. 0 ≤ ( a − b ) 2 0 \leq (a-b)^2 ⟹ 2 a b ≤ a 2 + b 2 \implies 2ab \leq a^2+b^2 So, ( a + b ) 2 = a 2 + b 2 + 2 a b ≤ a 2 + b 2 + a 2 + b 2 (a+b)^2=a^2+b^2+2ab \leq a^2+b^2+a^2+b^2 ⟹ ( a + b ) 2 ≤ 2 ( a 2 + b 2 ) \implies (a+b)^2 \leq 2(a^2+b^2) ⟹ a + b ≤ 2 c \implies a+b \leq \sqrt{2}c Here, the equality happens when a = b a=b .

Now, we have to find k k such that the following inequality is satisfied. a + b ≤ k c a+b \leq \sqrt{k}c k ≥ 2 k \geq 2 satisifies the above inequality. So, m i n ( k ) = 2 min(k) = 2

1 Helpful 1 Interesting 2 Brilliant 0 Confused

He/She asked for min(k) value. You found out max value. 😐

Debasish Roy - 1 year, 1 month ago

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I calculated the max value of m m , not k k . I added a few lines for further clarity. We were asked to find the minimum k k such that the inequality a + b ≤ k c a+b \leq \sqrt{k}c holds. As we can see, k k can be anything greater than 2 2 . So, the minimum value of k k is 2 2 .

Atomsky Jahid - 1 year, 1 month ago

We shall claim that the answer is 2.

Now, let us attempt to prove it.

Let a a be the length of one leg of the triangle, b b be the length of the other leg of the triangle, and a 2 + b 2 \sqrt{a^2+b^2} be the length of the hypotenuse. Hence, we have to prove: a + b ≤ 2 a 2 + 2 b 2 a+b\leq \sqrt{2a^2+2b^2} ⇔ a 2 + b 2 + 2 a b ≤ 2 a 2 + 2 b 2 \Leftrightarrow a^2+b^2+2ab\leq2a^2+2b^2 ⇔ a 2 + b 2 − 2 a b ≥ 0 \Leftrightarrow a^2+b^2-2ab\geq0 ⇔ ( a − b ) 2 ≥ 0 \Leftrightarrow (a-b)^2\geq0 This is obviously true, hence 2 is a possible value.

Obviously 1 is impossible, since triangle inequality is violated.

Thus, the smallest value of k k is 2.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

k need not be an integer, so the last part of your proof is invalid. However, it is easy to see that when a=b, a+b=root(2)c, so clearly no smaller k is possible.

Joe Mansley - 2 years, 9 months ago

I don't know what you guys are doing so I tried a more simplified approach: Set a=3 and b=4 (3-4-5 triangle). Then we'll get 1.96 <= k. So the answer might be 2. Turns out luck is on my side :))

0 Helpful 0 Interesting 0 Brilliant 0 Confused

One way to solve it is to maximize the hypotenuse. For a given total length of two legs, hypotenuse will be greatest when the two legs are equal.

Atomsky Jahid - 4 years, 12 months ago
Akash Shukla
Jun 6, 2016

a + b ≤ k ∗ c a+b≤\sqrt k*c ,

Squaring on both the sides,

a 2 + b 2 + 2 a b ≤ k ∗ c 2 a^2+b^2+2ab ≤ k*c^2

2 a b ≤ ( k − 1 ) c 2 2ab≤(k-1)c^2

k − 1 ≥ 2 a ∗ b c 2 k-1 ≥ 2\dfrac{a*b}{c^2}

k − 1 ≥ 2 s i n ( A ) ∗ s i n ( B ) s i n ( C ) 2 k-1 ≥2\dfrac{sin(A)*sin(B)}{{sin(C)}^2}

k − 1 ≥ 2 s i n ( A ) ∗ c o s ( A ) 1 k-1 ≥2\dfrac{sin(A)*cos(A)}{1} ... ( A = π 2 − B , s i n ( c ) = s i n ( π 2 ) = 1 ) ( A=\dfrac{\pi}{2}-B, sin(c) = sin(\dfrac{\pi}{2}) = 1) ,

k − 1 ≥ s i n ( 2 A ) k-1 ≥sin(2A) ,

k ≥ 1 + s i n ( 2 A ) k ≥ 1+sin(2A)

If my method is right, then min(k) will depend on either of angle A A or B B

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Since A is in between 0 to pi/2, 0< sin(2A) <=1; How would you conclude?

Tarikul Islam - 1 year, 1 month ago

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