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Official Solution:


On a plane, there are four distinct points A, B, C, and D. If triangle ABC is a right-angled triangle and BD is one of its heights, what is the product of the length of AB and BC?

Since the points A, B, C, and D are distinct and BD is a height of the right-angled triangle ABC, we know that point B must be the right angle and AC must be the hypotenuse. This means that BD is the height from right angle B to hypotenuse AC. So, the problem is asking us to find the product of the two non-hypotenuse sides AB and BC.

(1) AB = 6. Clearly insufficient.

(2) The two sides of triangle ABC that are not the hypotenuse have a product of 24.

As discussed above, these two sides are AB and BC. Therefore, we can directly find the value of AB*BC to be 24. Sufficient.


Answer: B
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Official Solution:


Since all points are distinct and BD is a height then B must be a right angle and AC must be a hypotenuse (so BD is a height from right angle B to the hypotenuse AC). Question thus asks about the product of non-hypotenuse sides AB and BC.

(1) AB = 6. Clearly insufficient.

(2) The product of the non-hypotenuse sides is equal to 24 \(\rightarrow\) directly gives us the value of AB*BC. Sufficient.


Answer: B

How do you know that B is a right angle? Couldn't we also have BC or BA as a hypotenuse with BD still indicating the height of the triangle?

Thx in advance for your help.
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first option says AB =6
doesnt it imply that the sides are 6,8,10?
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I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I am surprised if A,B, C, and D are distinct points and ABC is right triangle, right angled at B, then how come A , and D are not the same point. AB is the height . I am damn confused.
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Hi Bunuel,

Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous.
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Hi Bunuel,

Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous.

If A is a right angle, so if BA is perpendicular to CA, then the height from B to CA will be BA, making A and D to coincide, which is not possible since we are told that A, B, C, and D are distinct points on a plane. The same if C is a right angle.
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I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I am surprised if A,B, C, and D are distinct points and ABC is right triangle, right angled at B, then how come A , and D are not the same point. AB is the height . I am damn confused.

Each triangle can have three different bases and perpendicular to each base, three different heights. In the given description of triangle ABC, angle B being right angle, if you take BA or BC as height, then D coincides with A or C. But the question also says all four points are distinct. Hence we need to take AC as base and BD as height. Thats why this problem is in 700-800 difficulty level. Hope this helps.
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Bunuel
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Hi Bunuel,

Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous.

If A is a right angle, so if BA is perpendicular to CA, then the height from B to CA will be BA, making A and D to coincide, which is not possible since we are told that A, B, C, and D are distinct points on a plane. The same if C is a right angle.


How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...
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Hi Bunuel,

Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous.

If A is a right angle, so if BA is perpendicular to CA, then the height from B to CA will be BA, making A and D to coincide, which is not possible since we are told that A, B, C, and D are distinct points on a plane. The same if C is a right angle.


How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...

The figure is NOT right. We are given that BD is a height of the triangle. The height is a perpendicular dropped from one of the vertices to the opposite side.
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How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...[/quote]

The figure is NOT right. We are given that BD is a height of the triangle. The height is a perpendicular dropped from one of the vertices to the opposite side.[/quote]

-----

Hmm...I guess my concept of height wasn't clear. So, height of a triangle is ALWAYS the altitude of the triangle, inside the triangle.. and that altitude changes depending on which side is the 'base'.
Thanks!
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nk18967
How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...

The figure is NOT right. We are given that BD is a height of the triangle. The height is a perpendicular dropped from one of the vertices to the opposite side.[/quote]

-----

Hmm...I guess my concept of height wasn't clear. So, height of a triangle is ALWAYS the altitude of the triangle, inside the triangle.. and that altitude changes depending on which side is the 'base'.
Thanks![/quote]

It's not necessary to be inside a triangle:
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I think this is a poor-quality question and I don't agree with the explanation. Solution is incorrect. Answer should be (E). There is no way to prove that AB or BC are both legs, or alternatively that one is a leg and the other a hypotenuse. The height BD can lie outside the triangle contrary to the discussion posted in the forum.
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I think this is a poor-quality question and I don't agree with the explanation. Solution is incorrect. Answer should be (E). There is no way to prove that AB or BC are both legs, or alternatively that one is a leg and the other a hypotenuse. The height BD can lie outside the triangle contrary to the discussion posted in the forum.

That's not correct. The height in a right triangle is either one of the legs or the perpendicular from right angle to the hypotenuse. Thus the height of a right triangle cannot lie outside the triangle. Moreover, since A, B, C, and D are distinct points then BD cannot be the height from non-right angle because in this case it would coincide with one of the legs.
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In the second statement, it says the product of 2 non hypotenuse sides is 24. Then in that case, it can have multiple answers such as 2*12, 3*8, 4*6. With help of statement 1, we can say that AB*AC is 6*4 only. Hence, my doubt is, why cant the answer be C ?
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In the second statement, it says the product of 2 non hypotenuse sides is 24. Then in that case, it can have multiple answers such as 2*12, 3*8, 4*6. With help of statement 1, we can say that AB*AC is 6*4 only. Hence, my doubt is, why cant the answer be C ?

The question asks to find the value of AB*BC. (2) says that AB*BC = 24. We have our answer! Does it matter whether it's 1*24, 1/2*48, 500*24/500, or anything else?
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Hi Bunuel.

Statement 2 says that the product of non hypotenuse sides is 24. Why are we not considering BD here?

Can among AB, BC or BD, any of the two sides be non hypotenuse side ?

Happy new Year.

Regards
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sandysilva
Hi Bunuel.

Statement 2 says that the product of non hypotenuse sides is 24. Why are we not considering BD here?

Can among AB, BC or BD, any of the two sides be non hypotenuse side ?

Happy new Year.

Regards

(2) says: The product of the non-hypotenuse sides of triangle ABC is equal to 24. In triangle ABC, AC is hypotenuse, so non-hypotenuse sides are AB and BC. I think this is shown/explained on previous two pages several times.
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