# Tag Archives: area

## Inscribed Triangle Challenge

@MathCeyhun posed an interesting geometry problem yesterday.

Even more interesting is that, as @MathCeyhun noted in a subsequent tweet, none of the posted solutions is correct.  There have been a few posted answers, but no solutions, so I thought I’d give it a try.

OBSERVATIONS

• The perpendicular bisector of each chord was given, and perpendicular bisectors of chords always lie on radii of the circle.
• If r is the radius of the circle, then the lengths of the extensions of the perpendicular bisectors are r-1, r-2, and r-3.
• Nothing given guarantees anything special about the triangle, so I assumed it was scalene.  I called the side lengths 2x, 2y, and 2z to simplify the bisection labels.
• Adding the bisector extensions, the radii to the vertices, and segment names and labels gave me this.

[Irrelevant to this problem, but I just realized by looking at this image that every triangle can be dissected into three isosceles triangles with congruent sides and a shared vertex point at the triangle’s circumcenter.  Pretty.]

SETTING UP MY SOLUTION

Each isosceles triangle is bisected by the perpendicular bisector of its base from which I extracted three relationships from the Pythagorean theorem.

$x^2+(r-1)^2=r^2 \longrightarrow x^2=(2r-1) \cdot 1$

$y^2+(r-2)^2=r^2 \longrightarrow y^2=(2r-2) \cdot 2$

$z^2+(r-3)^2=r^2 \longrightarrow z^2=(2r-3) \cdot 3$

[The relationship between the half-sides, the extension of the radius beyond the triangle, and the radius is another pretty pattern.]

That gives 3 equations in 4 variables.  I needed one more to solve….

The area of $\Delta ABC$ can be expressed two ways:  as the sum of the areas of the isosceles triangles, and using Heron’s formula.  From the areas of the isosceles triangles,

$Area( \Delta ABC) = \frac{1}{2}(2x)(r-1) + \frac{1}{2}(2y)(r-2) + \frac{1}{2}(2z)(r-3)$
$Area( \Delta ABC) = x \cdot (r-1) + y \cdot (r-2) + z \cdot (r-3)$

The sides of $\Delta ABC$ are 2x, 2y, and 2z, so its semiperimeter is x+y+z and Heron’s formula gives its area as

$Area( \Delta ABC) = \sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}$.

The area of a given triangle is unique, so the two different area expressions are equivalent, giving a fourth equation.

SOLVING A SYSTEM & ANSWERING THE QUESTION

With four equations in four variables, I had a system of equations.  The algebra was messy, so I invoked my CAS to crunch it for me.

The question asked for the area of the triangle, so I just substituted my values back into the area formulas.

And 17.186… is clearly not one of the choices in the original problem.

A PLEA…

Recognizing the perpendicular bisectors, seeing all the right triangles, and connecting the multiple ways to describe the area of a triangle made this solution reasonably easy to find with the help of my computer algebra system (CAS), but I know the background algebra is, at best, cumbersome.  I hope there’s a more elegant solution, but I don’t see it.  Can anyone offer a suggestion?

Either way … this is definitely becoming a challenge problem for my Quantitative Reasoning class this coming week!

## Squares and Octagons, A compilation

My last post detailed my much-too-long trigonometric proof of why the octagon formed by connecting the midpoints and vertices of the edges of a square into an 8-pointed star is always 1/6 of the area of the original square.

My proof used trigonometry, and responses to the post on Twitter  and on my ‘blog showed many cool variations.  Dave Radcliffe thought it would be cool to have a compilation of all of the different approaches.  I offer that here in the order they were shared with me.

Method 1:  My use of trigonometry in a square.  See my original post.

Method 2:  Using medians in a rectangle from Tatiana Yudovina, a colleague at Hawken School.

Below, the Area(axb rectangle) = ab = 16 blue triangles, and
Area(octagon) = 4 blue triangles – 2 red deltas..

Now look at the two green, similar triangles.  They are similar with ratio 1/2, making

Area(red delta) = $\displaystyle \frac{b}{4} \cdot \frac{a}{6} = \frac{ab}{24}$, and

Area(blue triangle) = $\displaystyle \frac{1}{16} ab$

So, Area(octagon) = $\displaystyle 2 \frac{ab}{24}-4\frac {ab}{16}=\frac{1}{6}ab$.

QED

Method 3:  Using differences in triangle areas in a square (but easily extended to rectangles)from @Five_Triangles (‘blog here).

Full solution here.

Method 4:  Very clever shorter solution using triangle area similarity in a square also from @Five_Triangles (‘blog here).

Full second solution here.

Method 5:  Great option Using dilated kitesfrom Dave Radcliffe posting as @daveinstpaul.

Full pdf and proof here.

Method 6:  Use fact that triangle medians trisect each other from Mike Lawler posting as @mikeandallie.

Tweet of solution here.

Method 7:  Use a coordinate proof on a specific square from Steve Ingrassia, a colleague at Hawken School.  Not a quick proof like some of the geometric solutions, but it’s definitely different than the others.

If students know the formula for finding the area of any polygon using its coordinates, then they can prove this result very simply with nothing more than simple algebra 1 techniques.   No trig is required.

The area of polygon with vertices (in either clockwise or counterclockwise order, starting at any vertex) of $(x_1, y_1)$, $(x_2, y_2)$, …, $(x_n, y_n)$ is

$\displaystyle Area = \left| \frac{(x_1y_2-x_2y_1)+(x_2y_3-x_3y_2)+...+(x_{n-1}y_n-x_ny_{n-1})}{2} \right|$

Use a 2×2 square situated with vertices at (0,0), (0,2), (2,2), and (2,0).  Construct segments connecting each vertex with the midpoints of the sides of the square, and find the equations of the associated lines.

• L1 (connecting (0,0) and (2,1):    y = x/2
• L2 (connecting (0,0) and (1,2):   y=2x
• L3 (connecting (0,1) and (2,0):  y= -x/2 + 1
• L4 (connecting (0,1) and (2,2):  y= x/2 + 1
• L5 (connecting (0,2) and (1,0):  y = -2x + 2
• L6 (connecting (0,2) and (2,1):  y= -x/2 + 2
• L7 (connecting (1,2) and (2,0):  y = -2x + 4
• L8 (connecting (2,2) and (1,0):  y = 2x – 2

The 8 vertices of the octagon come at pairwise intersections of some of these lines, which can be found with simple substitution:

• Vertex 1 is at the intersection of L1 and L3:   (1, 1/2)
• Vertex 2 is at the intersection of L3 and L5:  (2/3, 2/3)
• Vertex 3 is at the intersection of L2 and L5:  (1/2, 1)
• Vertex 4 is at the intersection of L2 and L4:  (2/3, 4/3)
• Vertex 5 is at the intersection of L4 and L6:  (1, 3/2)
• Vertex 6 is at the intersection of L6 and L7:  (4/3, 4/3)
• Vertex 7 is at the intersection of L7 and L8:  (3/2, 1)
• Vertex 8 is at the intersection of L1 and L8:  (4/3, 2/3)

Using the coordinates of these 8 vertices in the formula for the area of the octagon, gives

$\displaystyle \frac{ \left| 1/3 +1/3+0+(-1/3)+(-2/3)+(-1/3)+0 \right|}{2} = \frac{2}{3}$

Since the area of the original square was 4, the area of the octagon is exactly 1/6th of the area of the square.

## Two Squares, Two Triangles, and some Circles

Here’s another fun twist on another fun problem from the Five Triangles ‘blog.  A month ago, this was posted.

What I find cool about so many of the Five Triangles problems is that most permit multiple solutions.  I also like that several Five Triangles problems initially appear to not have enough information.  This one is no different until you consider the implications of the squares.

I’ve identified three unique ways to approach this problem.  I’d love to hear if any of you see any others.  Here are my solutions in the order I saw them.  The third is the shortest, but all offer unique insights.

Method 1: Law of Cosines

This solution goes far beyond the intended middle school focus of the problem, but it is what I saw first.  Sometimes, knowing more gives you additional insights.

Because DEF is a line and EF is a diagonal of a square, I know $m\angle CEF=45^{\circ}$, and therefore $m\angle CED=135^{\circ}$.  $\Delta CEF$ is a 45-45-90 triangle with hypotenuse 6, so its leg, CE has measure $\frac{6}{\sqrt{2}}=3\sqrt{2}$.  Knowing two sides and an angle in $\Delta DEC$ means I could apply the Law of Cosines.

$DC^2 = 4^2 + (3\sqrt{2})^2 - 2\cdot (3\sqrt{2}) \cdot \cos(135^{\circ})=58$

Because I’m looking for the area of ABCD,  and that is equivalent to $DC^2$, I don’t need to solve for the length of DC to know the area I seek is 58.

Method 2: Use Technology

I doubt many would want to solve using this approach, but if you don’t see (or know) trigonometry, you could build a solution from scratch if you are fluent with dynamic geometry software (GeoGebra, TI-Nspire, GSP).  My comfort with this made finding the solution via construction pretty straight-forward.

1. Construct segment EF with fixed length 6.
2. Build square CEGF with diagonal EF.  (This can be done several ways.  I was in a transformations mood, so I rotated EF $90^{\circ}$ to get the other endpoints.)
3. Draw line EF  and then circle with radius 4 through point E.
4. Mark point D as the intersection of circle and line EF outside CEGF .
5. Draw a segment through points and C.  (The square of the length of CD is the answer, but I decided to go one more step.)
6. Construct square ABCD with sides congruent to CD.  (Again, there are several ways to do this.  I left my construction marks visible in my construction below.)
7. Compute the area of ABCD.

Here is my final GeoGebra construction.

Method 3: The Pythagorean Theorem

Sometimes, changing a problem can make it much easier to solve.

As soon as I saw the problem, I forwarded it to some colleagues at my school.  Tatiana wrote back with a quick solution.  In the original image, draw diagonal, CG, of square CEGF. Because the diagonals of a square perpendicularly bisect each other, that creates right $\Delta DHC$ with legs 3 and 7.  That means the square of the hypotenuse of $\Delta DHC$ (and therefore the area of the square) can be found via the Pythagorean Theorem.

$DC^2 = 7^2+3^2 = 58$

Method 4: Coordinate Geometry

OK, I said three solutions, and perhaps this approach is completely redundant given the Pythagorean Theorem in the last approach, but you could also find a solution using coordinate geometry.

Because the diagonals of a square are perpendicular, you could construct ECFG with its center at the origin.  I placed point C at (0,3) and point E at (3,0).  That means point D is at (7,0), making the solution to the problem the square of the length of the segment from (0,3) to (7,0).  Obviously, that can be done with the Pythagorean Theorem, but in the image below, I computed number i in the upper left corner of this GeoGebra window as the square of the length of that segment.

Fun.

## Finding area

I follow the Five Triangles ‘blog for cool math problems.  A recent one proved particularly nice.

At first I wasn’t sure this situation was invariant.  I didn’t see how fixing three triangle areas guaranteed a fixed quadrilateral area.  Not seeing an immediate general solution approach, I reasoned that if there was a solution, it worked for multiple overall configurations.  If it worked in general, then it must also work for any particular case I chose, so I made the cevians perpendicular.  That made each of the given area triangles right.  I modeled that by constructing the overall triangle with the cevian intersection at the origin and the legs of the given area triangles along the coordinate axes.

There are many ways to do this, but I reasoned that if there was a single answer, then any one of them would work.  A right triangle with legs of length 8 and 5 would have area 20.  Constructing that triangle in GeoGebra fixed the lengths of the legs of the other two triangles and the hypotenuses of the area 8 & 15 triangles intersected at a Quadrant II point.  Here’s my construction.

I  overlayed a polygon to create the quadrilateral and measured its area directly.  For fun, I also wrote algebraic equations for lines CB and DA, found the coordinates of point F by solving the 2×2 linear system, used that to derive the area of $\Delta BDF$, and determined the area of the quadrilateral from that.

While I realized that this approach was just a single case of the given problem, it absolutely convinced me that the solution was unique.  Once the area 20 triangle was defined (whether or not the triangle was right), a side and the area of each of the other two given triangles is known.  That meant the heights of the triangles would be determined and thereby the location of the quadrilateral’s fourth vertex.  So, I knew without a doubt that the unknown area was $27 cm^2$, but I didn’t know a general solution.

Chronology of the General Solution

While I worked more on the problem, I also pitched it to my Twitter network and asked a colleague at my school, Tatiana Yudovina, if she was interested in the problem.  Next is Tatiana’s initial solution, followed by my generic Geogebra construction, and a much shorter solution Tatiana created.  My conclusion takes the problem to a more generic state and raises some potential extensions.

Tatiana’s First Solution:

Leveraging the fact that triangles with the same base have equivalent height and area ratios, she created a system of equations that solved to eventually determine the quadrilateral’s area.

My Generic GeoGebra Solution:

While Tatiana was working on her algebraic answer, I was creating  a dynamic version on GeoGebra.  I built the area 20 triangle by first drawing a segment AB and measuring its length, a.  That meant the height of this triangle, h, was given by $\frac{1}{2} a \cdot h =20\longrightarrow h=\frac{40}{a}$. Then I constructed a perpendicular line to AB and used the “Segment with Fixed Length” tool and defined the length using the generic length of h as defined above to create segment AC.  This worked because GeoGebra defined the length of AB as a variable as shown below.

I used the “Compass” tool to create a circle with radius AC through the perpendicular line created earlier. Point D is the intersection of the circle and the normal line.  I then constructed a perpendicular to AD through D and placed a random point E on this new line.  Point E was the requisite height above AB to guarantee that $\Delta ABE$ always had area 20 which I confirmed by drawing the triangle and computing its area.

I hid AC, the circle, and both normals.  Segment AB was a completely independent object, and point E was free to move along the second “height” normal.  I measured AE and repeated the previous construction to create the area 15 triangle. Because BE was part of a cevian, I drew line BE to determine point J on the normal defining the final vertex of the area 15 triangle.

Again, I hid all of my constructions and repeated the process to create the final vertex, K, of the area 8 triangle off side BE of the area 20 triangle.  Extending segments AJ and BK defined point L, the final vertex of the quadrilateral.  Laying a quadrilateral in the figure let me compute its area.  Moving points A, B, and E around the screen and seeing the areas remain fixed is pretty compelling evidence that the quadrilateral’s area is always 27, and Tatiana’s proof showed why.  You can play with my final construction on GeoGebra Tube here.

Then Tatiana emailed me a much shorter proof.

Tatiana’s Short Solution:

Reversing the logic of her first solution, Tatiana reasoned that equivalent-altitude triangles had equal base and area ratios.

And the sum of X and Y gave the quadrilateral’s area.

Conclusion:

This problem was entertaining both in the solution and the multiple ways we found it.  Creating the dynamic construction gave  insights into the critical features of the problem.

Here are some potential extensions I developed for this problem.  I haven’t fully explored any of them yet, hoping some of my geometry students this year might take up the exploration challenge.  I’d love to hear if any of my readers have any further suggestions.

1. It might be interesting to create an even more dynamic construction with the areas of the three given triangles defined by sliders.
2. Can the quadrilateral’s area be expressed as a closed-form function of the areas of the three given triangles.
3. What happens on the boundaries of this problem?  That is, what happens if one of the side triangles was a degenerate with area 0? What would happen to the quadrilateral? Would would be the corresponding affect on the area formula from extension 2?
4. Extending 3 even further, if both given side triangles were degenerates with area 0, it seems that the area formula from extension 2 should collapse to the area of the final given non-zero triangle, but does it?

Thanks again, Five Triangles, for another great problem!

## Non-Calculus approach to Invariable Calculus Project

I shared my posts (here and here) on the Invariable Calculus Project in the AP Calculus Community.  Gary Litvin posted a response within the Community offering there a great non-calculus alternative solution to the original problem of the area of the triangle formed by the x- and y-axes and any Quadrant I tangent line to $\displaystyle \frac{1}{x}$.  Here’s a paraphrase of Gary’s approach.

Let $\displaystyle \frac{x}{a} + \frac{y}{b}=1$ be any Quadrant I tangent line to $\displaystyle y=\frac{1}{x}$.  (In case you don’t recognize it, this tangent equation uses the intercept form of a line–a is the x-intercept and b is the y-intercept.)  Because the line intersects the parabola in a single point, we can find that point by solving the system of equations defined by the two equations.  Substituting for y gives

$\displaystyle \frac{x}{a} + \frac{\frac{1}{x}}{b}=1$.

This is equivalent to $x^2 - a \cdot x+\frac{a}{b}=0$, a quadratic.  We could determine the value of x using the quadratic formula.  Because there is only one solution to this equation (there is only one point of intersection, the point of tangency), the discriminant must be zero.  That means

$\displaystyle (-a)^2 - 4\cdot \left( \frac{a}{b} \right)=0$

which can be rearranged to give $ab=4$ ($a=0$ is extraneous).  Therefore, the area of the triangle formed by the tangent line to $\displaystyle y=\frac{1}{x}$ and the coordinate axes is $\displaystyle Area=\frac{1}{2} ab=2$ no matter what the point of tangency.

Shiny.

## Invariable Calculus Project II

As Rocky hinted in his comment to my last post, $\displaystyle f(x)=\frac{k}{x}$  also has the constant area property.  Following a lead from Cohen, et al’s Student Research Projects in Calculus, I discovered long ago that with nothing more than knowledge of the product rule, the quotient and/or chain rule, and a healthy dose of some patient algebra manipulations, students could actually determine all functions which have the property that right triangles formed by the x– and y-axes and tangent lines to those functions have constant area.  This morning, I discovered a nice CAS approach that makes the problem accessible to far more students.

This makes another great project for calculus students who’ve just learned algebraic rules for differentiation:

Determine an equation for any twice-differentiable function–$y=g(x)$–with the property that all tangent lines to g in Quadrant I, along with the x- and y-axes, form triangles of constant area.

(Very minor) HINT:  At some point in solving this problem, you’ll need to make and use some assumptions about the values of $a, g(a), g'(a)$, and $g''(a)$.

SOLUTION ALERT!  Don’t read further if you want to solve this problem for yourself.

Assumptions:  Let $(a,g(a))$ be any arbitrary point on $y=g(x)$ in Quadrant I.  This makes $a>0$ and $g(a)>0$.  I also know $g'(a)<0$ because otherwise both of the x– and y-intercepts of the tangent line would not be positive, making the triangle’s area negative.  Finally, if $g''(a)=0$, then g would be a linear function, and there would be only one triangle.  To keep the problem interesting, I’m going to assume $g''(a)\ne 0$.

Setting up:  We no longer have a specific function, so everything must be in generalities.  A generalized equation for a tangent line to any function $y=g(x)$ at $x=a$ is

$y-g(a)=g'(a)\cdot (x-a)$.

From here, the generalized x-intercept is $\displaystyle a-\frac{g(a)}{g'(a)}$, and the y-intercept is $g(a)-a\cdot g'(a)$.  [Side note, the x-intercept is also the same form used in Newton’s Method for root approximations, a connection I’ll make later in the term when I’m teaching AP Calculus.]  Combining the generalized intercepts, I can write a generic area formula.

$\displaystyle Area = \frac{1}{2} \cdot \left( a - \frac{g(a)}{g'(a)} \right) \cdot \left( g(a) - a \cdot g'(a) \right)$

Differentiating and Cleaning Up:  Whenever I’ve used this problem in the past, my students and I have always used paper & pencil techniques.  While I’m quite comfortable with my algebraic manipulation skills, the significant majority of my students struggle with this part of the assignment.  As I was writing this post today, I finally had a technology insight that I should have years ago, given my long interest in CAS.  So, I’ll show the next portion in two different approaches, first traditional, and then via CAS.

APPROACH 1 – TRADITIONAL PAPER & PENCIL:  Trying to keep some brevity in this expanding post, I skip a few algebra steps below while providing some guiding explanations.  Finding a common denominator in the Area equation and recognizing a common factor leads to

$\displaystyle Area(a) = - \frac{1}{2} \cdot \frac{\left( a \cdot g'(a) - g(a) \right) ^2}{g'(a)}$ .

Applying the quotient rule with respect to a gives

$\frac{d(Area(a))}{da} = -\frac{1}{2} \cdot \frac{g'(a)\cdot 2(a\cdot g'(a)-g(a))(1\cdot g'(a)+a\cdot g''(a) - g'(a))- (a\cdot g'(a)-g(a))^2\cdot g''(a)}{(g'(a))^2}$ .

Remember that I seek functions whose tangent lines create constant area triangles, so $\displaystyle \frac{d(Area(a))}{da} = 0$.  Using this on the left and canceling some terms on the right gives

$\displaystyle 0 = -\frac{1}{2} \cdot \frac{2a\cdot g'(a)\cdot g''(a)(a\cdot g'(a)-g(a))-(a\cdot g'(a)-g(a))^2 \cdot g''(a)}{(g'(a))^2}$ .

Pulling out common factors and cleaning up a little more turns this into a completely factored form.

$\displaystyle 0 = - \frac{1}{2} \cdot \frac{(a\cdot g'(a)-g(a))\cdot g''(a)\cdot (a\cdot g'(a)+g(a))}{(g'(a))^2}$

APPROACH 2 – CAS:  Because the algebra was going to be complicated enough to make viewing on a handheld calculator very difficult, I used my TI-nSpire CAS Computer software to tackle the problem.  In the image below, I defined the area function in line 1 and differentiated with respect to a in line 2.  Equating the final equation to 0 gives the last algebraic line above from APPROACH 1.

Notice that I did not need to define a differentiation technique or to manipulate the factoring.  The results on both lines automatically accomplish the factoring I worked so carefully to establish earlier in APPROACH 1.  This is a beautiful example of what I see as a central benefit of CAS:  Keeping users focused on the mathematics of the problem situation.

Some students might actually be curious about how the challenge of differentiating line 1 could end up as relatively “clean” as the result in line 2.  GOOD!  CAS also inspires creative thinking.

Other than the differentiation step, everything else in APPROACH 1 was simple algebra.  Complicated, perhaps, but simple.  In fact, I don’t think it’s mathematics at all; it’s algebraic arithmetic.  I’m not disparaging the work or the approach, but I see mathematics as pattern recognition and big thinking.  I think CAS is completely justified in this problem.

Applying the Zero Product Property:  Our initial assumptions clear the denominator because $g'(a)<0$.  Because $g''(a)\ne 0$, I can eliminate that term, too.  With a and $g(a)$ both positive and $g'(a)<0$, the $(a\cdot g'(a)-g(a))$ term must be negative and therefore can be eliminated.  That drops the initially complicated differential equation to

$\displaystyle 0 = a\cdot g'(a)+g(a)$.

Finally–the Solution:  Depending on how much your students know, this last equation can be solved three different ways:  A) recognizing differentiation rules, B) solving a separable differentiable equation, or C) using a CAS solver.  I typically assign this problem so early in a calculus course that they have no idea what a differential equation is, making  the first approach the only available technique.  But this is also a great problem to introduce after learning about separable DEs.

APPROACH A:  If you look carefully, you can recognize the right side as the result of the product rule applied to $a\cdot g(a)$.  (In my experience, most students need some time, encouragement, and occasionally some hints to “see” this.)  Because the product rule result equals zero, the original expression must have equalled a constant.  That means $a\cdot g(a) = C$ for any constant, C.  Solving gives $\displaystyle g(a)=\frac{C}{a}$.  That means Rocky’s suggested family of functions at the top of this post, $\displaystyle f(x)=\frac{k}{x}$ not only produces triangles of constant area, it’s the only family of functions that does!  Very cool!

APPROACH B:  Rewriting the result of the Zero Product Property simplification using xs and ys gives $\displaystyle 0=x\cdot \frac{dy}{dx} +y$.  The variables can be rearranged to give $\displaystyle -\frac{dx}{x}=\frac{dy}{y}$.  Integration gives $-ln(x)+ln(C)=ln(y)$ for any random constant, $ln(C)$.  Logarithm properties lead to $y=\displaystyle \frac{C}{x}$, as before.

APPROACH C:  While I like the pattern recognition insights from the previous two approaches, the solution can also be found using a CAS.

On the TI-nSpire, c1 represents any random constant, so the DE solver again gives the same results.

Conclusion:  No matter what approaches you take, this problem shows that the only functions that have the property of their tangent lines producing constant area triangles.

Shiny.

## Invariable Calculus Project

Here’s one of my favorite calculus projects.  I initially discovered it over 20 years ago in Cohen, et al’s superb Student Research Projects in Calculus.

For $x>0$, what is true about every triangle formed by the x- and y-axes and any tangent line to $\displaystyle y=\frac{1}{x}$ ?  Prove thy claim.

I’d love to say nothing more than that, but I usually don’t.  The problem sounds vague in its statement, but is pretty simple to solve.  The hidden property is a delightful surprise.  I encourage you to try it out for yourself before reading further.

I just assigned the problem to one of my classes of seniors.  The class is a one-semester introduction to calculus for primarily students who’ve never been in honors and largely aren’t enamored by mathematics.  Most take the class to get an introduction to statistics (fall) and calculus (spring) before likely taking a course in one of these two in college and–for most–never taking another math course.  With that background in mind, I’ve probably scaffolded this iteration of the problem more than I should.  Here’s the assignment I gave them this week.

WARNING!  Partial Solution Alert!  Don’t read further if you want to solve the problem for yourself.

I typically use this project early in my introduction to derivatives and walk students through a little review and data gathering to help them discover the surprising hidden property.  While I don’t expect my students to do this, my default approach to geometric-type problems is to use a dynamic geometry package.  The animation below shows what happens when I varied the point of tangency while tracking the base, height, and area of the resulting triangle.

Well, I hope that animation screams something.  The x– and y-intercepts are the base and height, respectively, of a right triangle.  While those intercepts obviously vary as the point of tangency changes, the area of the triangle always seems to be 4.  It never changes!  If you’ve any geometry sense, something like that just shouldn’t happen.  So, is this a universal property, or is my animation misleading or limited in some way?  That’s a good question, and it requires proof.  Can you prove this apparent property about tangent lines to $\displaystyle y=\frac{1}{x}$?

FINAL SOLUTION ALERT!  Don’t read further if you want to prove this property for yourself.

For $\displaystyle f(x)=\frac{1}{x}$, $\displaystyle\frac{d}{dx}\left(f(x)\right)=\frac{-1}{x^2}$, so an equation for the tangent line to f at any point $x=a$ is

$\displaystyle \left(y-\frac{1}{a}\right)=\frac{-1}{a^2}\left(x-a\right)$.

The x-intercept of this generic line is $\left(2a,0\right)$, and its y-intercept is $\displaystyle \left(0,\frac{2}{a}\right)$.  Therefore, the area of the triangle formed by the x-and y-axes and the tangent line to f at any point $x=a$ is

$\displaystyle Area=\frac{1}{2}\cdot base\cdot height = \frac{1}{2}\cdot 2a\cdot\frac{2}{a}=2$.

Cool!  The triangle’s area is always 2, completely independent of the point of tangency!

EXTENSION:

Are there any other functions that have a similar property, or is $\displaystyle y=\frac{1}{x}$ alone in the mathematical universe for having constant area triangles?  Well, that’s a problem for another post.