Problem Of the Week Returns!

Much trickier, but I think I have it. For simplicity of typing, in what follows let x = the measure of angle ABC, let y = the measure of angle ACB, let z=the measure of angle BAC, and let w = the measure of angle ABD.

Now, we know that since AB = AD, w is also equal to the measure of angle ADB. From triangle ABD, we know that 2w + z = 180. Therefore z = 180 - 2w. From triangle ABC, we get x + y + z = 180. Substituting, we get 180 - 2w + x + y = 180, which simplifies algebraically to x + y - 2w = 0. We know from the problem statement that x-y = 30 or alternatively x = 30 +y. Therefore, 30 + 2y - 2w = 0. Rearranging gives 2w - 2y = 30, which gives w - y = 15.

Now, we have x-y = 30 and w-y = 15. Subtract the second of these equations from the first any you get x-w = 15. But (refresh yourself on the definitions I gave for x and w if you need to), x - w is the measure of angle CBD, which is what we were asked to find. Therefore the measure of angle CBD is 15.

As mathematicians, we are of course bothered by the undue specificity of the statement of the problem. We thus want to generalize the problem. The 30 is clearly too specific, we'll allow for other differences and just let the difference equal d. And we don't want to limit ourselves to a particular measurement system, we might want degrees, or grads, or radians, or some other manner or measuring angles. So we'll generalize to sum of interior angles in a plane triangle as T, be it 180, 200, pi, or whatever.

Now, in whatever system we are using, angle ACB measures as a. Thus angle ABC equals a+d. So the sum of those two angles is 2a+d, and angle BAC equals T-(2*a +d).

Triangle ABD is an isosceles triangle with line AD being the base, so angles ABD and ADB are equal. As angle BAD equals T-(2*a+d), the sum of ABD and ADB equals2*a+d, and each of the two equal base angles equals a+d/2.

Angle ABC equals a+d, and angle ABD equals a+d/2, so angle CBD equals d/2

Which, since in the problem d equals 30, angle CBD equals 15.

Just one question, are the angles all to be considered inner angles, or are we talking orientation? Because with orientation, one of the angles ABC or ACB is an outer angle. i never could remember if one goes clockwise or counterclockwise in this case.

Cthulhu,

Jason has made clear in the past that the POTW's are intended for a general audience, not specifically for mathematics majors. In that vein, the problems are intended to be straightforward, not tricky. In this vein, I believe it's safe to say that the angles in this problem were intended to all be regarding as interior angles, with measures in the range of 0 to 180 degrees.

You can make the algebra simpler. Let a= the angle of the base of the isosceles triangle, b = the angle we are trying to find, and c = ACB, the angle on the far right.

a is supplementary to the obtuse angle BDC, so b+c-a=0. We are given a+b-c=30. Add the two equations together to get 2b=30, so b=15.

I just spent an hour trying to solve the wrong version of the problem, so I wonder if anyone can help me with a slight modification.

Instead of AB = AD, what if AB = BD? Is the problem still solvable?

Jason,

I think your problem is unsolvable. Let's assume that a solution is possible and work toward it. Since AB=BD, angle BAC = angle BDA, call this value a. Then angle ABD = 180 - 2a. Let c be equal to angle ACB, and let b be the angle we are looking for, CBD. Now, angle ABC is 180-2a+b. We are given that 180-2a+b-c = 30.

Now consider angle BDC. It is clear that BDC is supplementary to angle BDA, which implies that 180-a is equal to this angle. It is also clear that the sum of this angle + b + c is also 180. Therefore 180-a+b+c=180, which implies that -a+b+c = 0 or equivalently a-b-c = 0.

Now we have two equations: 180-2a+b+c = 30 and a-b-c=0. Add those two equations. That gives 180-a = 30. This gives a = 150 degrees. However, a is, by construction the measure of the equal angles in an isosceles triangle. It is not possible for a to have a value of 150 degrees since there are two angles in the same triangle that have that value. Therefore, your version of the problem has no valid solution.

Uh Oh. I really wish there was an edit feature here. Scratch the above. I missed a negative sign that completely messes up everything. I will continue to work on your version, Jason.

Thanks, Sean! I appreciate it.

I found (I think) through trial and error that if angle BAC = 70, the triangle exists without contradiction, and in fact turns out to be isosceles with BAC = ABC. Filling in the rest of the angles, DBC = 30. I couldn't reverse engineer this into a proof, though.

Alas, Jason, you will never find your solution via proof because your solution is not unique. We can find constraints for possible solutions, but no unique solution exists. Let a = angle BAC = angle ADB. Let c = angle ACB. Let b = angle CDB, which is the one we are looking for. We also know that angle ABC - c = 30.

Thus, from triangle ABC, we get a+30+c+c = 180 or 2c+a = 150. We know a60 or c>30. We also get 180-2a = 30+c-b (two different expressions for angle ABD). That gives
2a+c-b = 150. Since a=b+c (exterior angle of triangle BCD), c = a-b. Substituting, 2a + a - b - b =150 or 3a-2b =150. That gives 3/2a - 75 for b. Thus a>50, for if not, 3/2a - 75 is negative, and b cannot be negative. Since 50<a<90, from this last formula, 0<b<60.

Second sentence, second paragraph should be "We know a30." Note that from a>50, and 2c+a=150, we also get 2c<100 or c<50. So then our constraints are:

50<a<90
0<b<60 and
30<c<50.

Sean is right. I had different names for the angles and came up with the same results:

a=isosceles base angle
b= isosceles vertex angle
x=angle we're looking for
c=angle at C

Then, as a varies, we have:

b=180-2a (this sets the upper limit for a at 90)
c=75-a/2
x=3a/2-75 (this sets the lower limit for a at 50)

Looking at the triangle ABD, let the angle at vertex A = a, that at B = b, and that at D = c. Call the angle we are looking for d.

Switching to triangle ABC, we can see the angle at B is b+d, so the angle at C is c-d.

The difference between angle ABC and angle BCA is thus (b-c)+2d.

In the original problem, b = c, so the difference is 2d.

In the modified problem, this is no longer the case. Here a=c, and b = 180-2*a. So the given difference is 180-2a-a +2d, or 180-3a+2d. Since the difference is 30, then it is 150 -3a+2d, or d = (3a/2 - 75).

Since the modified problem does not appear to constrain a to a single value, it appears there is not a single numerical solution.

One more note: the two problems become the same when a = 60 degrees so d = 15.