FInd the side lengths for an acute triangle
The side lengths of a triangle are n , n + 1 and n + 2 , where n is an integer. What is the smallest value of n such that the triangle is acute?
Details and assumptions
A triangle is acute if all of its angle are strictly less than 9 0 ∘ .
The answer is 4.
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6 solutions
please tell me what is wrong if we take n=2
Clearly n + 2 > n + 1 > n . Now, for this triangle to be acute, we must have that the sum of the squares of the smaller two sides be greater than the square of the larger side. Or
n 2 + ( n + 1 ) 2 > ( n + 2 ) 2 ⇒ 2 n 2 + 2 n + 1 > n 2 + 4 n + 4 ⇒ ( n − 3 ) ( n + 1 ) > 0
The above quadratic inequality is satisfied when x < − 1 or when x > 3 . Since x is positive, its smallest possible value is 4 .
why must the some of the squares of smaller two sides be greater than the square of the larger side. Do you have a proor? Something is reminding Pythagoras théorem,the sum of the small squares is equal to the square of hypothenuse in right triangle, so if the square of the bigger side is smaller than the sum , the correspondint angle is smaller than half pie; but why?
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Draw a right triangle. Try to increase/decrease the side of a hypotenuse.
First, you'll get a 2 + b 2 = c 2 .
c decreases when it becomes an acute triangle.
a 2 + b 2 > c 2 for an acute triangle.
c increases when it becomes an obtuse triangle.
a 2 + b 2 < c 2 for an obtuse triangle.
The proof for this is basically a generalization of the Pythagorean Theorem and can be easily proven by drawing the altitude of a triangle and then writing the square a side length in terms of the sum of squares of other side lengths.
For all the angles to be acute, the largest angle must be acute. The largest angle is opposite to largest side. Therefore cosine of angle opposite to largest side must be positive. Applying cosine rule
cosineA={(N)^2+(N+1)^2-(N+2)^2}/2(N+1)(N) >0
On solving we get ,
N>3 and N<-1(which is absurd)
Hence minimum value of N=4 since at N=3 cosineA is 0 i.e. angle A is 90 degrees
Let x=n+1. implies n=x-1, and n+2= x+1.
For all angles to be acute, sum of squares of two smaller sides must be greater than square on the longest side.
∴
(
x
−
1
)
2
+
x
2
>
(
x
+
1
)
2
⟹
x
2
−
2
x
+
1
+
x
2
>
x
2
+
2
x
+
1
∴
x
2
−
4
x
>
0
.
Gives 0>0 or x>4. That is n+1>4, n>3. since n is an integer, n=4.
A triangle is acute if c squared is less than a squared plus b squared, where c is the longest side. Plugging in n, n+1, and n+2, we simplify that to 2n+3 is less than n squared. 3 would make this equation be equal, so 4 is the next largest integer.
foe a triangle to be an acute ,2^ { largest side} < 2^{smaller side}+2^{smallest side}
therefore,
2^{n+2} <2^{n+1} +2^{n}
therefore , 2*2^{n}+2n+1 <2^{n}+4n +4
by solving this stuff we get n {min}=4
Obviously, we have n < n + 1 < n + 2 Therefore, n + 2 is the longest side. By the Pythagorean Inequality, we have n 2 + ( n + 1 ) 2 > ( n + 2 ) 2 Which can be expanded and simplified to get n 2 > 5 The smallest value of n is 3 , but n = 3 ⟹ n + 1 = 4 , n + 2 = 5 which is a Pythagorean Triple. Checking for n = 4 , we find 4 , 5 , 6 does not make a right triangle. Therefore, n = 4 .