New POTW Posted

The second Problem of the Week has now been posted. Also posted is an official solution to the first problem. Go have a look, and let me know what you think in the comments.

But don't get cocky. The problems get harder as we go along...

More like this

These are pretty easy. Do you mind having the answer and/or method posted. Suppose a student follows your blog?

I always make the first few problems fairly easy, because I want students in the lower-level classes to be able to participate as well. There are going to be ten problems during the term. The first five will be fairly simple, while the second five will be more challenging.

You can feel free to post solutions here. The kind of students who are motivated enough to participate in this are also the kind who enjoy solving the problems for themselves. Also, they are not receiving any kind of course credit for this.

Drop a perpendicular across L and M, touching the vertex of the 86 degree angle. The final angle x is computable via a series of supplementary and complementary angles starting at 133 and passing through the two right angles formed by the perpendicular at the 86 vertex.

Interestingly, the sum of the three labelled angles is 360. The construction above can be used to prove this, independent of the value of the two identified angles. I leave the derivation as an exercise for subsequent readers.

By Michael Kelsey (not verified) on 02 Feb 2015 #permalink

jrosenhouse wrote (February 2, 2015):
> The second Problem of the Week has now been posted. […]

> In the diagram […] lines $latex L$ and $latex M$ are parallel. Given the two angles as marked

… presumably with values of "degree" …

> find the measure of angle $latex x$.

Angle $latex x$ may have any value between (not including) 0 and 180 degrees.
Unless, of course, the vertex of the angle marked "86" were further constrained as being plane with respect to lines $latex L$ and $latex M$
(such that the line from vertex $latex x$ to vertex "86" could be extended to intersect line $latex L$).

> But don’t get cocky. The problems get harder as we go along…

Plainly so.

By Frank Wappler (not verified) on 03 Feb 2015 #permalink

Fairly trivial.
Extend the line segment that crosses the line M so that it cuts Line L to form a transversal.
The angles of the triangle (formed by the transversal, line L and the other non parallel line segment) are then the supplementary angle of 133 (ie 180-133), the supplementary angle of 86 (ie 180 - 86) and (180 - the other two angles) since the angles of a triangle add up to 180.
ie the acute angle between the extended line and line L is 39 degrees. Since L and M are parallel, the supplementary angle to x = the latter, so
x = 180 -39 = 141 degrees.

By Andrew Hobbs (not verified) on 03 Feb 2015 #permalink

Lots of potential solutions. Another would be: -
Taking your diagram as shown.
The two line segments intersect the two parallel lines at two points. Extend a line through the two intersection points. This forms a triangle with one of the angles being 86 degrees.
Angle x is then the sum of the acute angles between the added line and the line segment and the added line and line M.
The acute angle is equal to the angle at one of the vertices of the triangle.
The obtuse angle is equal to the obtuse angle between extended line and Line L, which in turn is equal to the sum of the angle at the 3rd vertex of the triangle, and the acute angle between the line segment and line L. This second angle is equal to the supplementary angle to the 133 given angle.
Since the sum of the angles of the formed triangle is 180 then
x= (180 - 86) + (180 - 133)
ie x= 141 degrees.

Of course this would be so much easier with a well labelled diagram.

By Andrew Hobbs (not verified) on 03 Feb 2015 #permalink

jrosenhouse wrote (#7, February 4, 2015):
> In what way is the diagram not well-labelled?

As pointed out above (#4) already:
The diagram at http://educ.jmu.edu/~rosenhjd/POTW/Spring15/POTW2S15.pdf would benefit from extending the line segment from "$latex x$" via "86" towards and beyond line "$latex L$",
in order to indicate their intersection (if there is one meant to be).
If so, if could even bear another "$latex x$" label.

By Frank Wappler (not verified) on 03 Feb 2015 #permalink

Frank,

From a strictly formal mathematical viewpoint, you are undoubtedly correct. However, in the previous POTW post, Jason made it clear that these problems are not meant to be formally mathematically correct. He stated the principle that the diagrams are not intended to be misleading. In that problem, we had a line that appeared to be, but was not explicitly defined to be, the radius of a circle. Jason made it clear that if it appears to be a radius, it is meant to be one. In similar vein, the vertex angle labeled as 86 degrees clearly appears to be an angle lying in the plane of the parallel lines, so I think it's safe to assume that it is in that plane, so there can only be one solution for x, namely 141 degrees as others have pointed out upthread.

Another solution is to draw a third parallel line through the 86-intersection. Then it is easy to see, that the three angles form a full circle, and thus have to be x = 360° - 86° - 133° = 141°.

Frank, when I first read your comment I honestly thought you were kidding around with me. You may as well object that I didn't stipulate that the numbers “133” and “86” were written in base ten. It would just be bizarre, given a diagram of the sort that appears in this problem, to stipulate that the vertex at 86 is in the same plane as the parallel lines. Of course it is. You would have to assume that the author of the problem was deliberately trying to be deceptive to interpret the diagram otherwise.

Jason Rosenhouse wrote (February 4, 2015):
> […] You may as well object that I didn’t stipulate that the numbers “133” and “86” were written in base ten.

Good point (?) … Nah, that might be at odds with the notation of the "Due" dates.

> You would have to assume that the author of the problem was deliberately trying to be deceptive

Or: that the author's splendid insincerity is presumed, to be appreciated.

By Frank Wappler (not verified) on 04 Feb 2015 #permalink

Or: that the author’s splendid insincerity is presumed,

Actually, the only reasonable conclusion is that your objection is completely unfounded.

The sum of the interior angles of a polygon equals 180 x (n - 2) where n is the number of sides. A line perpendicular to L and M to the left of x makes a 5 sided polygon with 4 of the 5 angles known. 540 - 90 - 90 - 133 - 86 = 141