Help me with this Geometry Question

[SophiaRivera007]

Hello all,

I was practicing some problems and I come across this question :

 

Is triangle ABC with sides a, b and c acute angled?

1. Triangle with sides a², b², c² has an area of 140 sq cms.
2. Median AD to side BC is equal to altitude AE to side BC.

Which one is the right answer? And How?

1. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
2. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
3. BOTH statements (1) and (2) TOGETHER are sufficient but NEITHER statement ALONE is sufficient to answer the question asked.
4. EACH statement ALONE is sufficient to answer the question asked.
5. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked; additional data specific to the problem is needed.

[Sriman Dutta]

From Fact 2 we know that the AE and AD are same, so altitude AE bisects BC. This implies that the triangle is either equilateral or isoceles.

[studiot]

I would be interested to see the context of this problem, since my first thoughts are that triangle AED is impossible in euclidian plane geometry.

Edited December 30, 2016 by studiot

[imatfaal]

I would be interested to see the context of this problem, since my first thoughts are that triangle AED is impossible in euclidian plane geometry.

 

As Sriman has pointed out the Part 2 means that the original triangle is either isosceles (with BC as base) or equilateral - thus AE is the same point as AD; this seems correct to me from a quick glance and scribble. This would also mean that there is no (as there can be no) triangle AED - exactly as you say; the internal angles of any triangle AED would have to be over 180 degrees as both angles AED and EDA would be right-angles.

OK - first Idea

 

1. If Median AD is the same as Altitude AE then line AD and line AE are the same and points D and E are at same point on line BC (side a).

2. If (1) is true then triangle is either isosceles with side a as base or equilateral.

3. Therefore length b=c and angle B=C

4. We know for a triangle to exist any two sides must be longer than the third

5. As we know triangle ABC exists then : a<b+c, b<c+a, c<a+b

6. As we are told that triangle made up of sides a^2,b^2,c^2 has an area (and thus exists as a triangle) we all know a^2<b^2+c^2, b^2<c^2+a^2, c^2<a^2+b^2

7. From 3 and 6 we know that a^2<2b^2

8. The Cosine rule states that a^2=b^2+c^2-2bcCosA

9. From 3 and 8 we know that a^2=b^2+b^2-2b^2CosA = 2b^2-2b^2CosA

10. Rearrange this to CosA= (a^2-2b^2)/(-2b^2)

11. From 7 and 10 we know that both numerator and denominator must be negative - thus CosA is positive

12. If Cos A is positive then A lies between 0 and 90 degrees

13. We know from 2 that angles B and C must be less than 90 degrees ( if equilateral then angles = 60 degrees; if isosceles then angles <90)

14. We can show that A is less than 90, and B and C are less than 90 - triangle is an acute triangle.

 

Just noticed that 5 is not necessary. Hope this is right

 

Both statements are necessary and sufficient to tell that ABC is acute triangle and neither statement on its own would be sufficient (3)

[michel123456]

 

If the above configurations are correct, it looks to me that the triangle can be either type 1, 2, 3.

I don't know how you can rule out triangle 2.

 

So IMO the answer is 5.

Edited December 30, 2016 by michel123456

[DrKrettin]

 

I don't know how you can rule out triangle 2.

 

.

 

In the previous post, it was shown that because of the first condition, the angle at the top must be less than 90 degrees. This rules out triangle 2

Edited December 30, 2016 by DrKrettin

[imatfaal]

IMG_20161230_171327.jpg

 

If the above configurations are correct, it looks to me that the triangle can be either type 1, 2, 3.

I don't know how you can rule out triangle 2.

 

So IMO the answer is 5.

 

Exactly as Dr Krettin said - condition 2 states that the area of triangle a^2 b^2 c^2 is something. This means we can use the fact that two sides of a triangle must be greater than the third, that sides AB and AC are equal, and the Cosine Rule to make the deduction that Cosine of A is positive. A triangle's angle can only have a positive Cosine if that angle is between 0 and 90 degrees (Cos is +ve 0-->90, -ve 90-->270, +ve 270-->360) . That rules out your triangle 2

 

With only Condition 1 we would be left with your three triangles - but the information that a^2 b^2 c^2 also forms a triangle gives us enough to rule out triangle 2

 

BTW your diagram is not how I would usually label in geometry - the side of a triangle (a,b,c) is named after the opposite angle (A,B,C) ; I wonder if this is a national / language thing.

 

http://mathworld.wolfram.com/Triangle.html

[studiot]

First I would like to thank imatfaal for reminding me to read the question properly.

 

I was creating some quite complicated equations, because I neglected to read condition 1 properly and thought the triangle area referred to the original triangle.

 

Stupid fool.

 

Anyway I was also trying to avoid imatfaal's solution and produce a purely geometric one (no trigonometry).

 

So here is the beginning of it.

 

Condition2 allows us to instantly determine that neither angle C nor angle B can be obtuse because the altitude AE is said to line on (intersect) BC as shown.

 

Euclid did produce a construction equivalent to the cosine rule and if anyone is interested they might like to replace the trigonometry with this.

 

Until that point here is my version using condition 1.

 

Edited December 30, 2016 by studiot

[michel123456]

 

Exactly as Dr Krettin said - condition 2 states that the area of triangle a^2 b^2 c^2 is something. This means we can use the fact that two sides of a triangle must be greater than the third, that sides AB and AC are equal, and the Cosine Rule to make the deduction that Cosine of A is positive. A triangle's angle can only have a positive Cosine if that angle is between 0 and 90 degrees (Cos is +ve 0-->90, -ve 90-->270, +ve 270-->360) . That rules out your triangle 2

 

With only Condition 1 we would be left with your three triangles - but the information that a^2 b^2 c^2 also forms a triangle gives us enough to rule out triangle 2

 

BTW your diagram is not how I would usually label in geometry - the side of a triangle (a,b,c) is named after the opposite angle (A,B,C) ; I wonder if this is a national / language thing.

 

http://mathworld.wolfram.com/Triangle.html

So you say that the triangle 2 cannot produce another triangle of sides a^2 b^2 c^2 and area 140 cm^2?

First I would like to thank imatfaal for reminding me to read the question properly.

 

I was creating some quite complicated equations, because I neglected to read condition 1 properly and thought the triangle area referred to the original triangle.

 

Stupid fool.

 

Anyway I was also trying to avoid imatfaal's solution and produce a purely geometric one (no trigonometry).

 

So here is the beginning of it.

 

Condition2 allows us to instantly and determine that both angles C and B cannot be obtuse because the altitude AE is said to line on (intersect) BC as shown.

 

Euclid did produce a construction equivalent to the cosine rule and if anyone is interested they might like to replace the trigonometry with this.

 

Until that point here is my version using condition 1.

 

acute1.jpg

Your sketch is different than mine. That is the root of the problem.

[studiot]

So you say that the triangle 2 cannot produce another triangle of sides a^2 b^2 c^2 and area 140 cm^2?

Your sketch is different than mine. That is the root of the problem.

 

In what way is it different?

 

Please look here for the geometry

 

https://en.wikipedia.org/wiki/Triangle_inequality

Edited December 30, 2016 by studiot

[imatfaal]

So you say that the triangle 2 cannot produce another triangle of sides a^2 b^2 c^2 and area 140 cm^2?

 

Not really; it is a proof of multiple parts.

 

1. All Condition 2 needs to tell us is that there exists a triangle with an area (ie not the silly degenerate triangle of zero area) - this means that side a^2 must be shorter than the sum of side b^2 plus side c^2. ie a^2<b^2+c^2

 

2. We know already that both the original triangle made of a b c and thus the second triangle of a^2 b^2 c^2 must be either equilateral or isosceles with side a / a^2 as the base.

 

3. This allows us to substitute b^2 for c^2 in the inequality in (1) to make a^2<b^2 + b^2 --> a^2 < 2b^2

 

4. Trigonometry tells us through the cosine rule that for a triangle XYZ with sides xyz that x^2 = y^2 + z^2 -2yzCos(X)

 

5. If we look at the cosine rule for angle A we get a^2 = b^2 +c^2 - 2bcCos(A).

 

6. We already know b=c so sub in to the equation in 5 and we get a^2 = b^2 + b^2 - 2.b.b.Cos(A) Simplified to a^2 = 2b^2 - 2b^2Cos(A)

 

7. If we rearrange equation in 6 we get Cos(A) = (a^2 - 2b^2) / (-2b^2)

 

8. The top of that fraction must be negative cos we know from (3) that a^2 is less than 2b^2

 

9. The bottom of that fraction must be negative cos it is simple and has a negative coefficient to b^2

 

10. If both top and bottom are negative the whole fraction is positive.

 

11. If Cosine is positive then angle A must be from 0<A<90 or 270<A<360 - for a triangle it must be 0<A<90

 

12. If angle A is between zero and ninety then your triangle B cannot be correct.

[michel123456]

 

Not really; it is a proof of multiple parts.

 

1. All Condition 2 needs to tell us is that there exists a triangle with an area (ie not the silly degenerate triangle of zero area) - this means that side a^2 must be shorter than the sum of side b^2 plus side c^2. ie a^2<b^2+c^2

 

2. We know already that both the original triangle made of a b c and thus the second triangle of a^2 b^2 c^2 must be either equilateral or isosceles with side a / a^2 as the base.

 

3. This allows us to substitute b^2 for c^2 in the inequality in (1) to make a^2<b^2 + b^2 --> a^2 < 2b^2

 

4. Trigonometry tells us through the cosine rule that for a triangle XYZ with sides xyz that x^2 = y^2 + z^2 -2yzCos(X)

 

5. If we look at the cosine rule for angle A we get a^2 = b^2 +c^2 - 2bcCos(A).

 

6. We already know b=c so sub in to the equation in 5 and we get a^2 = b^2 + b^2 - 2.b.b.Cos(A) Simplified to a^2 = 2b^2 - 2b^2Cos(A)

 

7. If we rearrange equation in 6 we get Cos(A) = (a^2 - 2b^2) / (-2b^2)

 

8. The top of that fraction must be negative cos we know from (3) that a^2 is less than 2b^2

 

9. The bottom of that fraction must be negative cos it is simple and has a negative coefficient to b^2

 

10. If both top and bottom are negative the whole fraction is positive.

 

11. If Cosine is positive then angle A must be from 0<A<90 or 270<A<360 - for a triangle it must be 0<A<90

 

12. If angle A is between zero and ninety then your triangle B cannot be correct.

Till point 3.

 

a^2 < 2b^2

 

The equation a^2=2b^2 is Pythagore and describes a right-angle triangle (with 2 equal sides b and hypotenuse a)

 

IOW a^2< 2b^2 describes automatically an obtuse* acute triangle and you are correct.

 

*edited.

Edited December 31, 2016 by michel123456

[imatfaal]

Till point 3.

 

a^2 < 2b^2

 

The equation a^2=2b^2 is Pythagore and describes a right-angle triangle (with 2 equal sides b and hypotenuse a)

 

IOW a^2< 2b^2 describes automatically an obtuse triangle and you are correct.

I think you are mixing ideas. a^2<2b^2 is nothing to do with pythagoras.

 

For any normall triangle side x is shorter than the addition of side y and z ie. x<y+z

 

As we know two of the sides are equal we can reword that as the base is shorter than the sum of the two other sides: x<2y

 

In this case we have been told that x = a^2 and y = b^2 so we get a^2 < 2b^2

[studiot]

 

 

1. Triangle with sides a², b², c² has an area of 140 sq cms

 

I wonder if the area of this triangle is red herring?

 

The area allows distinguishing between two possible cases of the sine rule, but we are using the cosine rule which is unambiguous.

The triangle inequality holds for all triangles therefore for all cases of the sine rule.

[michel123456]

I think you are mixing ideas. a^2<2b^2 is nothing to do with pythagoras.

 

For any normall triangle side x is shorter than the addition of side y and z ie. x<y+z

 

As we know two of the sides are equal we can reword that as the base is shorter than the sum of the two other sides: x<2y

 

In this case we have been told that x = a^2 and y = b^2 so we get a^2 < 2b^2

 

 

Yes, well understood. I edited my post above because I mixed the language.

 

 

a^2 < 2b^2 describes an acute triangle, a^2 = 2b^2 describes a right-angle triangle, and a^2 > 2b2 describes an obtuse triangle. Or is this a conjecture?

[imatfaal]

 

I wonder if the area of this triangle is red herring?

 

The area allows distinguishing between two possible cases of the sine rule, but we are using the cosine rule which is unambiguous.

The triangle inequality holds for all triangles therefore for all cases of the sine rule.

 

It is that there is an area - ie merely that the triangle actually exists and the triangle isn't the degenerate x=y+z . I think.

 

I think it is also a clever distraction - I was in the process of setting up huge and daunting simultaneous equations when I saw the similarity between the triangle inequality for the A'B'C' triangle to the cosine rule for the ABC triangle. Without the figure of 140 I would have been slower to do that

 

 

Yes, well understood. I edited my post above because I mixed the language.

 

 

a^2 < 2b^2 describes an acute triangle, a^2 = 2b^2 describes a right-angle triangle, and a^2 > 2b2 describes an obtuse triangle. Or is this a conjecture?

 

No. The first describes all isosceles and equilateral triangles, the second describes the degenerate triangle, and the third describes three sides which do not link up. Draw a line of length a^2 (just pick a length), then with a pair of compasses construct a triangle in which the sum of the other two sides is less than a^2 - it cannot be done

 

Remember that the lengths of the sides of the triangle in Condition 2 are a^2, b^2 and c^2 but the fact that these lengths are numerically the squares of the lengths of the main triangle does not mean that they do not have to obey the triangle inequality

[michel123456]

 

 

No. The first describes all isosceles and equilateral triangles, the second describes the degenerate triangle, and the third describes three sides which do not link up. Draw a line of length a^2 (just pick a length), then with a pair of compasses construct a triangle in which the sum of the other two sides is less than a^2 - it cannot be done

 

Remember that the lengths of the sides of the triangle in Condition 2 are a^2, b^2 and c^2 but the fact that these lengths are numerically the squares of the lengths of the main triangle does not mean that they do not have to obey the triangle inequality

I speak about a triangle of sides a, b, and c=b.

Not a triangle of sides a^2, b^2 and c^2=b^2

[imatfaal]

I speak about a triangle of sides a, b, and c=b.

Not a triangle of sides a^2, b^2 and c^2=b^2

 

Yes - but why? The question was about a triangle with sides a b c - but condition 2 of the question was concerning a triangle with sides a^2 b ^2 c^2.

 

basically the question allows you to say that if there is an isosceles triangle a b b and iff the triangle a^2 b^2 b^2 exists then all the angles of triangle a b b must be acute.

[michel123456]

 

(edited post. No comment)

Edited January 1, 2017 by michel123456

[michel123456]

Right.

What I say is the following:

The question was
Is triangle ABC with sides a, b and c acute angled?

• Triangle with sides a2, b2, c2 has an area of 140 sq cms.
• Median AD to side BC is equal to altitude AE to side BC.


Condition 2 says that the triangle ABC (1) with sides abc is an isosceles triangle IOW side b is equal to side c
. This condition alone does not help to tell if triangle 1 is acute.

Condition 1 is about another triangle, I will call triangle T.
We can insert this in condition 1 and say that b^2 =c^2 which means that the second triangle of sides a^2 b^2 c"2 is also isosceles.
IOW the sketch made by Studiotin post#8 doesn't correspond. In this sketch the altitude is not median.

So now we have 2 triangles that are both isosceles.

From your step 3 in your analysis, you get that a^2 < 2b^2 for triangle T
and we have the equivalent for triangle (1) that is to say a<2b
And we know that the values abc are the same in both triangles. IOW we can exchange statements about a & b from one triangle to another.

We can have a look at the special case of Pythagore because this is the triangle that separates acute from obtuse. In our case the isosceles triangle (1) is a right-angle triangle (90 degrees at A), which says that a^2=b^2+c^2 which reduces to a^2=2b^2
This contradicts the statement a^2 < 2b^2, IOW triangle t cannot be a right-angle triangle. Triangle t is acute.

Edited January 1, 2017 by michel123456

[studiot]

Michel please note you have reproduced the only correct part of my post.

 

This part did not involve the median in any way. It simply points out that if the altitude runs from A to BC ie if E lies between B and C as stated

 

Then angles B and C must be acute.

 

The second part of my post was in error since I mixed up the triangles with sides a, b and c and with sides a², b², and c².

[SophiaRivera007]

IMG_20161230_171327.jpg

 

If the above configurations are correct, it looks to me that the triangle can be either type 1, 2, 3.

I don't know how you can rule out triangle 2.

 

So IMO the answer is 5.

Thanks for easy solution Mr. michel123456

[michel123456]

Thanks for easy solution Mr. michel123456

No I must have been wrong then.

 

Here below the empirical way

 

 

All the black triangles have a surface of 140cm2 (not to scale, the image is extracted from an Autocad file). These are the T triangles.

 

The small red triangles have sides squared roots of the black ones. These are the t triangles.

 

The small t triangles reduce to a right-angle triangle as shown on the right side. As much it becomes flat, as much it gets close to a^2=2b^2.

Which makes me conclude I was correct in my earlier post #20

 

To be noted that Imatfaal was correct before me.

 

And you can call me Michel.

Edited January 2, 2017 by michel123456

[imatfaal]

Right.

 

What I say is the following:

 

The question was

Is triangle ABC with sides a, b and c acute angled?

 

• Triangle with sides a2, b2, c2 has an area of 140 sq cms.

• Median AD to side BC is equal to altitude AE to side BC.

 

 

Condition 2 says that the triangle ABC (1) with sides abc is an isosceles triangle IOW side b is equal to side c

.Screen Shot 01-01-17 at 01.28 PM.JPG This condition alone does not help to tell if triangle 1 is acute.

 

Condition 1 is about another triangle, I will call triangle T.

We can insert this in condition 1 and say that b^2 =c^2 which means that the second triangle of sides a^2 b^2 c"2 is also isosceles.

IOW the sketch made by Studiotin post#8 Screen Shot 01-01-17 at 01.34 PM.JPGdoesn't correspond. In this sketch the altitude is not median.

 

So now we have 2 triangles that are both isosceles.

 

From your step 3 in your analysis, you get that a^2 < 2b^2 for triangle T

and we have the equivalent for triangle (1) that is to say a<2b

And we know that the values abc are the same in both triangles. IOW we can exchange statements about a & b from one triangle to another.

 

We can have a look at the special case of Pythagore because this is the triangle that separates acute from obtuse. In our case the isosceles triangle (1) is a right-angle triangle (90 degrees at A), which says that a^2=b^2+c^2 which reduces to a^2=2b^2

This contradicts the statement a^2 < 2b^2, IOW triangle t cannot be a right-angle triangle. Triangle t is acute.

 

I see where you were going now. That is neat - and Studiot get's his non-trigonometrical answer.

[studiot]

I'd like to applaud the spirit of cooperation evident in this thread.