Archive for the ‘Geometry’ Category

I was perusing some old AP Calculus exams recently and ran across an interesting problem.  The free-response questions are an interesting bunch.  I won’t analyze or critique them too much except to say that they tend to be kind of the same, without much variety.

The question I was really drawn to presents the graph of a derivative function and asks a series of questions about the maximum/minimum values and points of inflection of the underlying function.  It says that if the graph below is f(x) and g(x)=\int_2^xf(x)\;dx, then etc, etc.

The graph of the derivative looks like this:

The test questions based on the graph aren’t all that interesting, but I got really interested in wanting to see the original function.  I suppose you can integrate the piecewise derivative graph and use the identified points to build a piecewise function, but I did this geometrically, since these are all triangles.  Really I was just interested in what the original function looked like – which will appear after the jump for those of you who want to think about this for a minute…


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One of my students gave me a problem last fall that was very interesting.

The problem is posed with the following diagram:


What is the measure of \angle CDA?  That is, what is the value of x?

I’ve given a similar problem in my trigonometry class for the past few years, except that version of the problem has a side length included and the triangle is not isosceles.

A pdf of this problem is linked below:


Working from my experience with the other version of this problem, I began to write in values for the various unlabeled angles in the diagram – if we label the intersection of \overline{AD} and \overline{BC} as K, then \angle CKD and \angle AKB are both 70^{\circ}, \angle CKA and \angle DKB are both 110^{\circ}, which makes \angle KCA 50^{\circ} and \angle ADB is 40^{\circ}.

I added in new variables and created a system of four equations with four unknowns, but it was a dependent system.

The solution for this problem that was devised by the student who gave it to me is after the jump…


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