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The three sides of triangle have length p, q and r, each an integer.

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The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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The three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?

(1) The perimeter of the triangle is an odd integer.
(2) If the triangle's area is doubled, the result is not an integer.


Please hit kudos if you liked the question :)
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 20 May 2017, 10:56
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poojamathur21 wrote:
Hi Bunuel,

Could you please share the solution for this? I'm very confused!


The three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?


(1) The perimeter of the triangle is an odd integer.

Given p + q + r = odd. This implies that either all three p, q and r are odd OR two of them are even and the third one is odd.

If all three p, q and r are odd, then \((side_1)^2 + (side_2)^2 = odd+odd=even\) and \((side_3)^2=odd\). Thus \((side_1)^2 + (side_2)^2 \neq (side_3)^2\). The triangle cannot be right angled.

If two of them are even and the third one is odd, then \((side_1)^2 + (side_2)^2 = even+even=even\) and \((side_3)^2=odd\) OR \((side_1)^2 + (side_2)^2 = even+odd=odd\) and \((side_3)^2=even\). In any case \((side_1)^2 + (side_2)^2 \neq (side_3)^2\). The triangle cannot be right angled.

So, we have a definite answer to the question: the triangle is NOT right angled. Sufficient.

From above we can derive that if the lengths of all three sides of a right triangle are integers (so if a right triangle is Pythagorean Triple triangle), then the perimeter of the triangle must be even.

(2) If the triangle's area is doubled, the result is not an integer.

If the triangle were right angled, the area would be \(\frac{(leg_1)*(leg_2)}{2}\) and double that would be \(2*\frac{(leg_1)*(leg_2)}{2}=(leg_1)*(leg_2)=integer*integer=integer\). Since we are told that double the area of the triangle is NOT an integer, then the triangle is NOT right angled. Sufficient.

Answer: D.

Hope it's clear.
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 22 Mar 2017, 12:45
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The Three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?

Two points to be known about a Right Angled Triangle

The height and the hypotenuse of Right Angled Triangle will always be Odd Integers
The Base of the Right Angled Triangle will always be an Even Integer

Statement 1) the perimeter of the triangle is an odd integer.
Based on the above, the perimeter of a Right Angled Triangle will always be Even. Hence the Triangle in question is not a right Angled Triangle. This statement is sufficient.

Statement 2) if the triangle's area is doubled, the result is not an integer.

The Area of a Right Angled Triangle is 1/2 * b*h. On doubling this it becomes b*h. Since all sides of a Right Angles Triangle are integers, b*h will also be an integer. So in this case, the Triangle in question is not a right Angled Triangle. This statement is sufficient.

Answer is D
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 22 Apr 2017, 09:13
quantumliner wrote:
The Three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?

Two points to be known about a Right Angled Triangle

The height and the hypotenuse of Right Angled Triangle will always be Odd Integers
The Base of the Right Angled Triangle will always be an Even Integer

Statement 1) the perimeter of the triangle is an odd integer.
Based on the above, the perimeter of a Right Angled Triangle will always be Even. Hence the Triangle in question is not a right Angled Triangle. This statement is sufficient.

Statement 2) if the triangle's area is doubled, the result is not an integer.

The Area of a Right Angled Triangle is 1/2 * b*h. On doubling this it becomes b*h. Since all sides of a Right Angles Triangle are integers, b*h will also be an integer. So in this case, the Triangle in question is not a right Angled Triangle. This statement is sufficient.

Answer is D



Consider : 6,8,10... These also form right angled triangle.
So,I think reasoning for St 1 may not be correct...
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 01 May 2017, 04:42
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nainy05 wrote:
quantumliner wrote:
The Three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?

Two points to be known about a Right Angled Triangle

The height and the hypotenuse of Right Angled Triangle will always be Odd Integers
The Base of the Right Angled Triangle will always be an Even Integer

Statement 1) the perimeter of the triangle is an odd integer.
Based on the above, the perimeter of a Right Angled Triangle will always be Even. Hence the Triangle in question is not a right Angled Triangle. This statement is sufficient.

Statement 2) if the triangle's area is doubled, the result is not an integer.

The Area of a Right Angled Triangle is 1/2 * b*h. On doubling this it becomes b*h. Since all sides of a Right Angles Triangle are integers, b*h will also be an integer. So in this case, the Triangle in question is not a right Angled Triangle. This statement is sufficient.

Answer is D



Consider : 6,8,10... These also form right angled triangle.
So,I think reasoning for St 1 may not be correct...









Is it safe to say that the perimeter of a right triangle is always even?
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 01 May 2017, 05:56
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satsurfs wrote:
nainy05 wrote:
quantumliner wrote:
The Three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?

Two points to be known about a Right Angled Triangle

The height and the hypotenuse of Right Angled Triangle will always be Odd Integers
The Base of the Right Angled Triangle will always be an Even Integer

Statement 1) the perimeter of the triangle is an odd integer.
Based on the above, the perimeter of a Right Angled Triangle will always be Even. Hence the Triangle in question is not a right Angled Triangle. This statement is sufficient.

Statement 2) if the triangle's area is doubled, the result is not an integer.

The Area of a Right Angled Triangle is 1/2 * b*h. On doubling this it becomes b*h. Since all sides of a Right Angles Triangle are integers, b*h will also be an integer. So in this case, the Triangle in question is not a right Angled Triangle. This statement is sufficient.

Answer is D



Consider : 6,8,10... These also form right angled triangle.
So,I think reasoning for St 1 may not be correct...


Is it safe to say that the perimeter of a right triangle is always even?


If the lengths of all three sides are integers (so if a right triangle is Pythagorean Triple triangle), then yes the perimeter will always be even.
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 20 May 2017, 10:24
Hi Bunuel,

Could you please share the solution for this? I'm very confused!
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 28 May 2017, 14:40
Bunuel wrote:
poojamathur21 wrote:
Hi Bunuel,

Could you please share the solution for this? I'm very confused!


The three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?


(1) The perimeter of the triangle is an odd integer.

Given p + q + r = odd. This implies that either all three p, q and r are odd OR two of them are even and the third one is odd.

If all three p, q and r are odd, then \((side_1)^2 + (side_2)^2 = odd+odd=even\) and \((side_3)^2=odd\). Thus \((side_1)^2 + (side_2)^2 \neq (side_3)^2\). The triangle cannot be right angled.

If two of them are even and the third one is odd, then \((side_1)^2 + (side_2)^2 = even+even=even\) and \((side_3)^2=odd\) OR \((side_1)^2 + (side_2)^2 = even+odd=odd\) and \((side_3)^2=even\). In any case \((side_1)^2 + (side_2)^2 \neq (side_3)^2\). The triangle cannot be right angled.

So, we have a definite answer to the question: the triangle is NOT right angled. Sufficient.

From above we can derive that if the lengths of all three sides of a right triangle are integers (so if a right triangle is Pythagorean Triple triangle), then the perimeter of the triangle must be even.

(2) If the triangle's area is doubled, the result is not an integer.

If the triangle were right angled, the area would be \(\frac{(leg_1)*(leg_2)}{2}\) and double that would be \(2*\frac{(leg_1)*(leg_2)}{2}=(leg_1)*(leg_2)=integer*integer=integer\). Since we are told that double the area of the triangle is NOT an integer, then the triangle is NOT right angled. Sufficient.

Answer: D.

Hope it's clear.


Bunuel my interpretation of your explanation is that if the pythagorean theorem cannot be validated then the triangle cannot be a right triangle- because the Pythagorean Theorem only applies to right triangles?
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 28 May 2017, 18:21
Bunuel wrote:
poojamathur21 wrote:
Hi Bunuel,

Could you please share the solution for this? I'm very confused!


The three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?


(1) The perimeter of the triangle is an odd integer.

Given p + q + r = odd. This implies that either all three p, q and r are odd OR two of them are even and the third one is odd.

If all three p, q and r are odd, then \((side_1)^2 + (side_2)^2 = odd+odd=even\) and \((side_3)^2=odd\). Thus \((side_1)^2 + (side_2)^2 \neq (side_3)^2\). The triangle cannot be right angled.

If two of them are even and the third one is odd, then \((side_1)^2 + (side_2)^2 = even+even=even\) and \((side_3)^2=odd\) OR \((side_1)^2 + (side_2)^2 = even+odd=odd\) and \((side_3)^2=even\). In any case \((side_1)^2 + (side_2)^2 \neq (side_3)^2\). The triangle cannot be right angled.

So, we have a definite answer to the question: the triangle is NOT right angled. Sufficient.

From above we can derive that if the lengths of all three sides of a right triangle are integers (so if a right triangle is Pythagorean Triple triangle), then the perimeter of the triangle must be even.

(2) If the triangle's area is doubled, the result is not an integer.

If the triangle were right angled, the area would be \(\frac{(leg_1)*(leg_2)}{2}\) and double that would be \(2*\frac{(leg_1)*(leg_2)}{2}=(leg_1)*(leg_2)=integer*integer=integer\). Since we are told that double the area of the triangle is NOT an integer, then the triangle is NOT right angled. Sufficient.

Answer: D.

Hope it's clear.


Hi Bunuel,

Can you please explain why the triangle is NOT right angled, if the area of the triangle is not an integer?
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 28 May 2017, 20:59
Nunuboy1994 wrote:
Bunuel wrote:
poojamathur21 wrote:
Hi Bunuel,

Could you please share the solution for this? I'm very confused!


The three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?


(1) The perimeter of the triangle is an odd integer.

Given p + q + r = odd. This implies that either all three p, q and r are odd OR two of them are even and the third one is odd.

If all three p, q and r are odd, then \((side_1)^2 + (side_2)^2 = odd+odd=even\) and \((side_3)^2=odd\). Thus \((side_1)^2 + (side_2)^2 \neq (side_3)^2\). The triangle cannot be right angled.

If two of them are even and the third one is odd, then \((side_1)^2 + (side_2)^2 = even+even=even\) and \((side_3)^2=odd\) OR \((side_1)^2 + (side_2)^2 = even+odd=odd\) and \((side_3)^2=even\). In any case \((side_1)^2 + (side_2)^2 \neq (side_3)^2\). The triangle cannot be right angled.

So, we have a definite answer to the question: the triangle is NOT right angled. Sufficient.

From above we can derive that if the lengths of all three sides of a right triangle are integers (so if a right triangle is Pythagorean Triple triangle), then the perimeter of the triangle must be even.

(2) If the triangle's area is doubled, the result is not an integer.

If the triangle were right angled, the area would be \(\frac{(leg_1)*(leg_2)}{2}\) and double that would be \(2*\frac{(leg_1)*(leg_2)}{2}=(leg_1)*(leg_2)=integer*integer=integer\). Since we are told that double the area of the triangle is NOT an integer, then the triangle is NOT right angled. Sufficient.

Answer: D.

Hope it's clear.


Bunuel my interpretation of your explanation is that if the pythagorean theorem cannot be validated then the triangle cannot be a right triangle- because the Pythagorean Theorem only applies to right triangles?


Yes. If a triangle is right angled , then \((side_1)^2 + (side_2)^2 = (side_3)^2\). The reverse is also true, if \((side_1)^2 + (side_2)^2 = (side_3)^2\), then a triangle is right angled.
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 28 May 2017, 21:01
adityapareshshah wrote:
Bunuel wrote:
poojamathur21 wrote:
Hi Bunuel,

Could you please share the solution for this? I'm very confused!


The three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?


(1) The perimeter of the triangle is an odd integer.

Given p + q + r = odd. This implies that either all three p, q and r are odd OR two of them are even and the third one is odd.

If all three p, q and r are odd, then \((side_1)^2 + (side_2)^2 = odd+odd=even\) and \((side_3)^2=odd\). Thus \((side_1)^2 + (side_2)^2 \neq (side_3)^2\). The triangle cannot be right angled.

If two of them are even and the third one is odd, then \((side_1)^2 + (side_2)^2 = even+even=even\) and \((side_3)^2=odd\) OR \((side_1)^2 + (side_2)^2 = even+odd=odd\) and \((side_3)^2=even\). In any case \((side_1)^2 + (side_2)^2 \neq (side_3)^2\). The triangle cannot be right angled.

So, we have a definite answer to the question: the triangle is NOT right angled. Sufficient.

From above we can derive that if the lengths of all three sides of a right triangle are integers (so if a right triangle is Pythagorean Triple triangle), then the perimeter of the triangle must be even.

(2) If the triangle's area is doubled, the result is not an integer.

If the triangle were right angled, the area would be \(\frac{(leg_1)*(leg_2)}{2}\) and double that would be \(2*\frac{(leg_1)*(leg_2)}{2}=(leg_1)*(leg_2)=integer*integer=integer\). Since we are told that double the area of the triangle is NOT an integer, then the triangle is NOT right angled. Sufficient.

Answer: D.

Hope it's clear.


Hi Bunuel,

Can you please explain why the triangle is NOT right angled, if the area of the triangle is not an integer?


I think this is explained. We know that the lengths of the sides of the triangle are integers. If the triangle were right angled, the area would be \(\frac{(leg_1)*(leg_2)}{2}\) and double that would be \(2*\frac{(leg_1)*(leg_2)}{2}=(leg_1)*(leg_2)=integer*integer=integer\). Since we are told that double the area of the triangle is NOT an integer, then the triangle is NOT right angled. Sufficient.
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 29 May 2017, 23:37
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giobas wrote:
The three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?

(1) The perimeter of the triangle is an odd integer.
(2) If the triangle's area is doubled, the result is not an integer.


Please hit kudos if you liked the question :)


Responding to a pm:

A few things about primitive pythagorean triplets (not multiples of other pythagorean triplets) a, b, c where c is the hypotenuse.
1. One of a and b is odd and the other is even.
2. c is odd.

(3, 4, 5), (5, 12, 13), (8, 15, 17) etc

The non-primitive triplets are made by multiplying each member of the primitive triplet by an integer n greater than 1. Depending on whether n is odd or even, the three sides will be (Odd, Even, Odd) or (Even, Even, Even).

Given:
p, q and r are all integers.

(1) The perimeter of the triangle is an odd integer.
The perimeter of a right triangle can never be odd.
Odd + Even + Odd = Even
Even + Even + Even = Even
Hence, the perimeter will be even in all cases.

(2) If the triangle's area is doubled, the result is not an integer.
If p, q and r are the sides of a right triangle such that r is the hypotenuse (could be any of p, q, r),
Area of this triangle = (1/2)*p*q
Double of area of this triangle = p*q
This has to be an integer.
But we know that this is not an integer. In that case, this triangle cannot be a right triangle.

If it is not a right triangle, double the area will be base*altitude. The altitude would not be an integer in this case.

Answer (D)
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 08 Jul 2017, 07:37
nainy05 wrote:
quantumliner wrote:
The Three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?

Two points to be known about a Right Angled Triangle

The height and the hypotenuse of Right Angled Triangle will always be Odd Integers
The Base of the Right Angled Triangle will always be an Even Integer

Statement 1) the perimeter of the triangle is an odd integer.
Based on the above, the perimeter of a Right Angled Triangle will always be Even. Hence the Triangle in question is not a right Angled Triangle. This statement is sufficient.

Statement 2) if the triangle's area is doubled, the result is not an integer.

The Area of a Right Angled Triangle is 1/2 * b*h. On doubling this it becomes b*h. Since all sides of a Right Angles Triangle are integers, b*h will also be an integer. So in this case, the Triangle in question is not a right Angled Triangle. This statement is sufficient.

Answer is D



Consider : 6,8,10... These also form right angled triangle.
So,I think reasoning for St 1 may not be correct...



6,8,10 in its simplest form will be 3,4,5.
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 22 Jul 2018, 01:07
The three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?

(1) The perimeter of the triangle is an odd integer.
(2) If the triangle's area is doubled, the result is not an integer.

Thought of pythagorean triplets first --> 3,4,5/ 12,13,5/ 8/10/6, 8/15/17 --> each of these have an even integer divisible by two, and each's perimeter is even cz in each case o + o + e is e (I did this by adding)

So Qx is asking us if triangle pqr is actually right angled or not.

St 1: triangle perimeter is odd. Based on quite a few pairs, perimeter is even. Which means, this triangle isn't a right triangle with integers. Sufficient.

St 2: If the triangle's area is doubled, the result is not an integer. --> if the double of something isn't divisible by 2 (I'm using formula 1/2 bh for area) so halfing it will also not be divisible by 2. Each pythagorean triplet right triangle has an even integer as its height, so this triangle in question isn't a right triangle. Sufficient

So answer is D.
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Re: The three sides of triangle have length p, q and r, each an integer.  [#permalink]

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New post 13 Jan 2019, 02:42
giobas wrote:
The three sides of triangle have length p, q and r, each an integer. Is this triangle right triangle?

(1) The perimeter of the triangle is an odd integer.
(2) If the triangle's area is doubled, the result is not an integer.


Please hit kudos if you liked the question :)


Since statement 1 can be easily evaluated using pythagorean triplets. Here is another way to evaluate statement 2:

Area of triangle is \(= 1/2*p*q*sin\alpha\), where \(\alpha\) is the angle between p & q. Now if we double this area, it will become \(p*q*sin\alpha\). The only thing you need to know is that \(sin 90 = 1\) and for any other angle the value of \(sin\) is non-integer. Since doubling the area makes it non integer and p and q are already integers => \(sin \alpha\) is non integer => Hence we know that the angle between p and q is not 90. So statement 2 is sufficient.
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Re: The three sides of triangle have length p, q and r, each an integer.   [#permalink] 13 Jan 2019, 02:42
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