[bonanova]

Consider an acute triangle ABC with angles α, β, γ as shown.

The figure is not necessarily to scale.

Let the bisector of angle BAC [α] intersect BC at P.

Let the bisector of angle ABC [β] intersect AC at Q.

Let AB + BP = AQ + QB.

Let α = 60^o.

Find β and γ.

[Guest]

is it 60 and 60...

[Guest]

beta = 80°

gamma = 40°

my proof is messy and lengthy though. I'd like to see the "simple" way to solve this.

[Guest]

Beta = 75 degrees

Gamma = 45 degrees

Am I right?

[HoustonHokie]

beta = 80°

gamma = 40°

my proof is messy and lengthy though. I'd like to see the "simple" way to solve this.

Like you, I have a long proof, but I got the same answer. Then I drew it out (as well as the other two solutions that have been posted), and confirmed we're right. I'm looking for that "magic bullet" answer - there always seems to be one

[Izzy]

Alpha = 60° = Given

AB + BP = AQ + QB = Given

a/h =30(cos)

Let's plug in random numbers of side lengths just because I can and it shouldn't affect much 'cos you can change that in a triangle and the angles can stay the same? ?

(I think we're meant to do that thing where we set up two equations and cancel stuff out to solve for more than one variable, but I have to leave in a few minutes...)

30(cos) =0.866

8.66/10 = 0.866

So, AB = 10 and AR (calling that bit in the middle R btw) = 8.66

(sin)ABR = o/h

(sin)ABR = 0.866/10

Huh. Man, I think I forgot how to use this stuff, it's been a while. Getting pretty wacky numbers. And... I have to leave for school.

Am I on the right track though?

[Guest]

Beta = 75 degrees

Gamma = 45 degrees

????

[HoustonHokie]

Beta = 75 degrees

Gamma = 45 degrees

????

I get AQ+QB < AB+BP with those angles.

[Guest]

Ok, using a tool for number calculations (like http://ostermiller.org/calc/triangle.html), we can easily validate a solution:

Let's just say AB = 1

For the solution beta = 80° and gamma = 40°

With Triangle ABQ, we get:

AQ = 0.65270

BQ = 0.87939

With triangle ABP, we get

BP = 0.53209

AP = 1.0480 )

Therefore, AQ+BQ = 1.53209 = BP + AB

This shows that these values are *a* valid solution (maybe not the only one, though)

For the solution beta = 75° and gamma = 35°

With ABQ, we get:

AQ = 0.61401

BQ = 0.87350

With ABP:

BP = 0.51764

Therefore, AQ+BQ = 1.48751 != BP + AB = 1.51764

This one is therefore wrong.

Ehe equilateral triangle one is easily disporven.

Now what we'd love is a nice 2-line proof ^^

[Prof. Templeton]

in extending line AB past point B to B¹ such that AB¹C is an equilateral triangle. Extend line AP to P¹ so that P¹ lies on B¹C. And the rest is complicated. I'll see what I can do if no one beats me to it.

[Guest]

Here's what I got. I hope it's right.

First off, I know from experience that the intersection of angle bisectors in a triangle marks the center of an inscribed circle in the triangle. Therefore, for the lines AP and BQ, if we call their intersection point O, then OP=OQ. LEt's call this r. Let's also call the angle AOQ theta. It's also clear that AOQ=BOP=theta. Finally, let's call angle ABC 2x, so that the half angle of B is x and the half angle of A is 30.

First, use the sin rule in traingle AOQ. sin30/r = sin(theta)/AQ. And in traingle BOP, sinx/r = sin(theta)/BP. Therefore AQsin(30)=BPsin(x), or sin(30)/BP = sin(x)/AQ.

Next, consider the traingle ABP compared to ABC. Since all their internal angles must add up to 180, it is easy to see that angle APB = gamma + 30. Similarly, angle AQB = gamma + x. Finally, use the sin rule on these two traingles. For triangle APB, sin(30)/BP = sin(gamma+30)/AB. And for triangle AQB, sin(x)/AQ = sin(gamma+x)/AB. Since we proved earlier that sin(30)/BP = sin(x)/AQ, it is easily seen that sin(gamma+30) = sin(gamma+x)

So, according to this solution, beta = 60 and gamma = 60 is the correct answer.

[Prof. Templeton]

Here's what I got. I hope it's right.

First off, I know from experience that the intersection of angle bisectors in a triangle marks the center of an inscribed circle in the triangle. Therefore, for the lines AP and BQ, if we call their intersection point O, then OP=OQ. LEt's call this r. Let's also call the angle AOQ theta. It's also clear that AOQ=BOP=theta. Finally, let's call angle ABC 2x, so that the half angle of B is x and the half angle of A is 30.

First, use the sin rule in traingle AOQ. sin30/r = sin(theta)/AQ. And in traingle BOP, sinx/r = sin(theta)/BP. Therefore AQsin(30)=BPsin(x), or sin(30)/BP = sin(x)/AQ.

Next, consider the traingle ABP compared to ABC. Since all their internal angles must add up to 180, it is easy to see that angle APB = gamma + 30. Similarly, angle AQB = gamma + x. Finally, use the sin rule on these two traingles. For triangle APB, sin(30)/BP = sin(gamma+30)/AB. And for triangle AQB, sin(x)/AQ = sin(gamma+x)/AB. Since we proved earlier that sin(30)/BP = sin(x)/AQ, it is easily seen that sin(gamma+30) = sin(gamma+x)

So, according to this solution, beta = 60 and gamma = 60 is the correct answer.

Does not necessarily cut angle BAC in half.

[Guest]

Does not necessarily cut angle BAC in half.

Line AP is the angle bisector of angle BAC. It said so in the problem.

[Prof. Templeton]

Line AP is the angle bisector of angle BAC. It said so in the problem.

So it does. Still, I don't think your solution is right.

[Guest]

So it does. Still, I don't think your solution is right.

I don't know if it satisfies AB + BP = AQ + QB. I'm not sure how to verify that. The one thing I'm concerned about is that the other answer given, beta = 80 and gamma = 40, doesn't satisfy sin(gamma+30) = sin(gamma+x), since sin(70) doesn't equal sin(80).

[HoustonHokie]

Here's my proof:

For the proof, I introduced two new angles, e and f, as follows:

e is the angle opposite side AB on triangle ABP.

f is the angle opposite side AB on triangle ABQ.

By law of sines, we know that:

BQ/sin(a) = AQ/sin(b/2) = AB/sin(f)

BP/sin(a/2) = AP/sin(b) = AB/sin(e)

We also know that:

a/2 + b + e = 180 Þ b + e = 150

a + b/2 + f = 180 Þ b/2 + f = 120

Finally, we are given that:

AQ + BQ = AB + BP

Substituting for AQ, BQ, & BP gives:

AB sin(b/2)__AB sin(a)______AB sin(a/2)

------------- + ----------- = AB + --------------

__sin(f)______sin(f)__________sin(e)

All terms on both sides of the equation include AB, so AB can drop out. Multiplying both sides by sin(e) sin(f) gives:

sin(e) [sin(b/2) + sin(a)] = sin(f) [sin(e) + sin(a/2)]

Substituting for a, e, and f gives:

sin(150 – b) [sin(b/2) + sin(60)] = sin(120 – b/2) [sin(150 – b) + sin(30)]

We now have one equation with one unknown: b.

Solving for b, we get b = 80 and therefore g = 40.

[Guest]

Very nice solution Hokie. Mine was a little different, but I think I wandered around useless directions a bit too much.

Artificial, your solution seems good, but the result is obviously wrong: in an equilateral triangle (60,60,60), a bissector would also be a height and a median. So if AB = 1, we'd have AQ = BP = 1/2. That would mean that BQ must be equal to 1, while it would actually be equal to sqrt(3)

The flaw in your argument is there:

First off, I know from experience that the intersection of angle bisectors in a triangle marks the center of an inscribed circle in the triangle. Therefore, for the lines AP and BQ, if we call their intersection point O, then OP=OQ.

Yes, bissectors interesct at the center of the inscribed circle O. This means that the shortest distance from O to the segment AC is the same as the shortest distance from O to the segment BC. However, these shortest distances are not OQ and OP, but rather segments that are perpendicular to AC and BC, respectively. Taking the assumption that these are the same is basically forcing your triangle to be equilateral

Edited February 4, 2009 by hellmake

[Guest]

Yes, bissectors interesct at the center of the inscribed circle O. This means that the shortest distance from O to the segment AC is the same as the shortest distance from O to the segment BC. However, these shortest distances are not OQ and OP, but rather segments that are perpendicular to AC and BC, respectively. Taking the assumption that these are the same is basically forcing your triangle to be equilateral

Ahhh, yes, I see it now. OQ and OP would only be the radii of the circle if they were perpendicular to AC and BC. And since that was my initial assumption, the rest of my solution was all wrong.

[Prof. Templeton]

Extend line AB to a point B' such that BP eguals B'P, this gives an isosoles triangle B'BP with base angles b/2. Since AP is an angle bisector, B'P and PC are equal and angle APB' equals angle APC. Proved congruent triangles (SAS), g = b/2. 120 - b = g = b/2. b=80 and g = 40.

[Guest]

I did it the way Hokie did but I got b=17.72 and g=107.28...

Will bonanova confirm any of the answers?

[Guest]

Extend line AB to a point B' such that BP eguals B'P, this gives an isosoles triangle B'BP with base angles b/2. Since AP is an angle bisector, B'P and PC are equal and angle APB' equals angle APC. Proved congruent triangles (SAS), g = b/2. 120 - b = g = b/2. b=80 and g = 40.

Your proof is really simple and leads to the correct answer, BUT

You've not used the data of AB+BP=AQ+BQ, if you agree this, then you used only the data of alfa=60⁰ .

This means that any acute triangle that's one angle is 60⁰, must always have two other angles as 80⁰ and 40⁰.

Maybe I've missed that you used more data other than a unique angle, if so, a clarification is requested.

[bonanova]

Extend line AB to a point B' such that BP eguals B'P, this gives an isosoles triangle B'BP with base angles b/2. Since AP is an angle bisector, B'P and PC are equal and angle APB' equals angle APC. Proved congruent triangles (SAS), g = b/2. 120 - b = g = b/2. b=80 and g = 40.

In your proof you say: "Since AP is an angle bisector, B'P and PC are equal ... "

First, you must show AB' = AC.

Which needs the invocation of AB+BP = AQ+QB; and that answers nobody's question.

Parenthetically,

Extend line AB to a point B' such that BP eguals B'P should read

Extend line AB to a point B' such that BP eguals B'B

But the drawing makes that clear.

[bonanova]

Will bonanova confirm any of the answers?

80^o and 40^o.

Quite by accident, that's nearly the way it appears in OP. [i used a triangle from a previous puzzle!]

HH derived the answer using the all-powerful law of sines.

Hellmake derived and verified the solution

PT gave a geometric proof for the answer, but a step might have been missing.

That is, B' needs to satisfy two conditions: BB' = BP [by construction] and PB' = PC [needs to be proved]

Proof of the 2nd condition [which needs AB+BP = AQ+QB] was, probably inadvertently, omitted.