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An exam for high school students had the following problem:

Let the point $E$ be the midpoint of the line segment $AD$ on the square $ABCD$. Then let a circle be determined by the points $E$, $B$ and $C$ as shown on the diagram. Which of the geometric figures has the greater perimeter, the square or the circle?

enter image description here

Of course, there are some ways to solve this problem. One method is as follows: assume the side lengths of the square is $1$, put everything somewhere on a Cartesian coordinate system, find the midpoint of the circle using the coordinates of $E$, $B$ and $C$, then find the radius of the circle, and finally use the radius to calculate the circle's circumference and compare it to the perimeter of the square.

The problem with that method is that ostensibly this problem is supposed to be very simple; it shouldn't require the student to know the formula for the midpoint of a circle given three coordinates. Therefore the question here is: does there exist a simple way to solve the problem without knowing any complicated geometric formulas?

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    $\begingroup$ All the answers so far involve computing the radius of the circle, and then essentially using the fact that $\pi < 16/5$. I wonder if there is a way to prove that $\pi < 16/5$ geometrically, or otherwise solve this problem without knowing the familiar fact that $\pi \approx 3.14$. $\endgroup$ – Phillip Andreae Mar 14 '15 at 1:49
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    $\begingroup$ There is no such thing as "require the student to know the formula for ..." in mathematics. This isn't science, anything in math that you don't know you can figure out. Your proposed approach really isn't complicated: long does not mean complicated. Short proofs are the complicated ones. $\endgroup$ – DanielV Mar 14 '15 at 2:02
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    $\begingroup$ @DanielV Of course one can just derive everything, but unfortunately things don't work like that in practice, especially if you are a high school student with limited time to solve each problem. :-) Therefore, for all intents and purposes the student would need to know that formula if they were to try the method I described. $\endgroup$ – Sid Mar 14 '15 at 2:23
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    $\begingroup$ Is the formula for the circumradius of a triangle $R=\frac{abc}{4S}$ known to your students, from the time when they studied Euclidean geometry? That one almost trivializes the problem. $\endgroup$ – Federico Poloni Mar 14 '15 at 11:38
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    $\begingroup$ @Sid probably because the title gives us a seemingly very simple comparison (see: is the dress blue or white), and then tells us that, "nope, it's hard". The person reading the title ends up taking it as a challenge. All of that without actually telling us anything about the real problem. Well done, perfect advertising :). $\endgroup$ – Millie Smith Mar 18 '15 at 18:23

24 Answers 24

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Perhaps the examiner intended the students to notice the square is determined by a $(3, 4, 5)$ triangle, because $3 + 5 = 4 + 4$ (!):

A circle and square

Consequently, as several others have noted, $$ \frac{\text{perimeter of the circle}}{\text{perimeter of the square}} = \frac{5 \cdot 2\pi}{4 \cdot 8} = \frac{\pi}{3.2} < 1. $$


For an approach less dependent on inspiration, taking the origin of the coordinate system at the center of the circle seems easier than placing the origin at the center of the square. Without loss of generality, assume the circle has unit radius:

Expressing the dimensions of the square

Equating the lengths of the horizontal and vertical sides of the square in this diagram, we read off $$ x + 1 = 2y\quad\text{(or $x = 2y - 1$).} $$ Invoking the Pythagorean theorem and substituting the preceding line, \begin{align*} 0 &= x^{2} + y^{2} - 1 \\ &= (2y - 1)^{2} + y^{2} - 1 \\ &= 5y^{2} - 4y \\ &= y(5y - 4). \end{align*} Clearly $y \neq 0$, so $y = 4/5$, $x = 3/5$, and we notice the Examiner's Favorite Triangle.

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    $\begingroup$ I like “Examiner's Favorite Triangle”, thanks! $\endgroup$ – MJD Mar 14 '15 at 17:19
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    $\begingroup$ This is a beautiful solution, so much so that it feels like the examiner started with the solution and worked backwards to the problem. Which isn't exactly fair to those taking the exam. (I certainly wouldn't have come up with it myself.) $\endgroup$ – fluffy Mar 14 '15 at 22:07
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    $\begingroup$ 'Notice the... (3,4,5) triangle'? Is there a quick observation that leads to that? Like the others, that seems to be an observation after all the calculations for the solution have been done. $\endgroup$ – Mitch Mar 15 '15 at 0:38
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    $\begingroup$ @Mitch - The statement $3 + 5 = 4 + 4$ definitely isn't sufficient on its own, else I could make up something like $3.2 + 5 = 4.1 + 4.1$. We need to combine this constraint with the right triangle constraint. After getting those two constraints, I think OP is suggesting you might guess that 3/4/5 meets the constraints. (I personally wouldn't have jumped to this, though perhaps it's easier to guess if you just learned about Pythagorean triples.) $\endgroup$ – blah Mar 15 '15 at 6:10
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    $\begingroup$ @user86418: nice... that's the super simple start. But you should add that to the answer to make it complete (i.e. where $x+1 = 2y$ comes from, both $x+1$ and $2y$ equal a side of the square). $\endgroup$ – Mitch Mar 16 '15 at 2:25
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Here's a proof that arrives at the ratio of circle diameter to square side using similar triangles rather than the Pythagorean theorem:

Perimeter comparison

Draw the diameter of the circle from $E$; let the other endpoint of that diameter be $F$ and let it cross the square at $G$. Then since the angle $\measuredangle EBF$ is a right angle, the triangles $\bigtriangleup EBG$ and $\bigtriangleup BFG$ are similar. Obviously $|EG|=2|BG|$, so $|BG|=2|FG|$ (or in other words $|FG|=\frac12|BG|$) and so $|EF|=|EG|+|FG|=2|BG|+\frac12|BG|=\frac52|BG|$. From here the proof proceeds just like the others: the perimeter of the square is $4|EG|=8|BG|$ and the circumference of the circle is $\pi|EF|=\frac{5\pi}{2}|BG|$. From $\pi\lt\frac{16}5=3.2$, we get $5\pi\lt16$ and $\frac{5\pi}{2}\lt 8$, so the perimeter of the square is longer.

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    $\begingroup$ I like this one because it follows a pattern (which is both standard and teachable) for solving elementary geometry: "Find similar triangles and expect everything else to follow". $\endgroup$ – Keith Mar 16 '15 at 2:52
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    $\begingroup$ @Keith Thank you - my main goal was to find a solution that one could actually derive and not just 'happen upon' (which, as amazing as they are, most of the Pythagorean-based solutions feel like to me), and I think this hits that mark. $\endgroup$ – Steven Stadnicki Mar 16 '15 at 3:26
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    $\begingroup$ @user86418 I use coordinates for a living, so I wholeheartedly understand that sentiment. :-) I really think the two solutions make a great complement for each other - though I'm still waiting for a 'clean' proof that $\pi\lt \frac{16}5$ (and I've been thinking of asking it as a separate question); Calum's is a nice one (and well deserving of upvotes) but uses too many pieces to (IMHO) really be clean. $\endgroup$ – Steven Stadnicki Mar 16 '15 at 16:14
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    $\begingroup$ The most natural approaches (which involve chopping off the corner or symmetrically chopping off a couple of chunks) fall just short; I spent far too much time chopping off symmetric pieces yesterday only to discover that that approach can only get to $\pi\lt24-12\sqrt{3}\approx3.215$ - so close, and yet... $\endgroup$ – Steven Stadnicki Mar 16 '15 at 16:14
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    $\begingroup$ Even simpler: by the Intersecting Chord Theorem, BG × GC = EG × GF, or 1 × 1 = 2 × GF, or GF = 1/2. Therefore the diameter = 2.5, C = πd ≈ 7.85, perimeter of square = 4 * 2 = 8. $\endgroup$ – Tebbe Mar 17 '15 at 2:19
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Regarding Abel's comment that he would like to see a less computational proof, consider the following. It is apparent that the length of a side of the square is $4$ small squares, and the perpendicular bisector of $EB$ crosses the horizontal centre line of the drawing $2.5$ small squares from $E$. Apologies for the inaccurate drawing.

enter image description here

More formally, if we call the centre of the square the origin:

EB has endpoints $(-2,0)$ and $(2,2)$ therefore it has midpoint $(0,1)$ and gradient $\frac12$.

The perpendicular bisector of EB has gradient $\frac{-1}{\mathrm{gradient\ of\ EB}}=-2$ therefore it crosses the $x$-axis at $(\frac12,0)$

Perimeter of ABCD = $16$

Perimeter of circle = $2 * \pi * 2.5 \approx 15.71$



(I liked this answer so much I wanted to illustrate step-by-step the process of finding the location of the centre of the circle:

First, the centre of the circle is defined by intersection of diameters; one is obviously the line straight across from $E$, and the other is generated by the perpendicular from the midpoint of $EB$:

enter image description here

Then we can locate the midpoint of $EB$ as being halfway across the square and one-quarter down from the top:

enter image description here

Then we can find another point on the perpendicular by turning the rectangle from E to the midpoint through a right angle:

enter image description here

And it's then obvious that the radius of the circle is $\frac{5}{8}$ the side of the square.

Joffan)

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    $\begingroup$ What's interesting is the way I seem to have stumbled across a "pythagorean triple." the centre of the circle is 5/2 from E, but is also sqrt( (3/2)^2 + (4/2)^2 ) = 5/2 from B. $\endgroup$ – Level River St Mar 14 '15 at 1:29
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    $\begingroup$ It's the famous $3,4,5$ Pythagorean triple. $\endgroup$ – David K Mar 14 '15 at 3:38
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    $\begingroup$ @DavidK indeed, but I wasn't expecting it to pop out like that. I knew the triangle made by the centre of the circle and EB was isoceles. So I knew the radius through B had to be rational, and there it was. I was thinking of a general method for finding triples, but in hindsight it's obvious that if you pick any intersection on a piece of squared paper and draw all the circles of integer radius around it, the circles will intersect with the grid at points corresponding to pythagorean triples. $\endgroup$ – Level River St Mar 14 '15 at 11:53
  • $\begingroup$ Alternatively, we can take any vector with integer x and y and form an isoceles triangle with the x axis and we will find a pythagorean triple. Plugging the gradient of the line into the double angle tangent formula is equivalent to another method of finding triples: take any x+yi with integer x and y and square it. Example: tangent formula 2*(0.5)/(1-0.5^2)=4/3; square of complex number (2+i)^2 = 3+4i. $\endgroup$ – Level River St Mar 14 '15 at 12:11
  • $\begingroup$ Why not subdivide tha square by eight right away? $\endgroup$ – Hagen von Eitzen Mar 16 '15 at 17:22
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i will let the side of the square be $2.$ the center of the circle is at the intersection of the perpendicular bisectors of the chords $BC$ and $EB$. let that point be $O$ and the foot of the perpendicular from $O$ be $F.$ now the triangle $OEF$ is similar to $EBA.$ therefore $$\frac{OE}{EB}= \frac{EF}{BA} \to \frac{OE}{\sqrt 5} = \frac{\sqrt 5/2}{2}\to OE = 5/4.$$

the circumference of the circle is $$2\pi \frac 5 4= 5\pi/2 = 7.85 $$ which is less than $8$ the circumference of the square. they are very close.

i would like to see a proof that is not as computational as this one.

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This isn't a full answer, just an observation to supplement the other answers (and celebrate Pi Day).

If you construct tangents at the 20 points $(0, \pm25)$, $(\pm25, 0)$, $(\pm15, \pm20)$, $(\pm20, \pm15)$, $(\pm7, \pm24)$, $(\pm24, \pm7)$ on the circle of radius 25 centred on the origin, they intersect at the vertices $(\pm\frac{25}{7}, \pm25)$, $(\pm25, \pm\frac{25}{7})$, $(\pm\frac{125}{11}, \pm\frac{250}{11})$, $(\pm\frac{250}{11}, \pm\frac{125}{11})$, $(\pm\frac{125}{7}, \pm\frac{125}{7})$ of an irregular circumscribing icosagon, whose perimeter is $8 \times \left(\frac{25}{7} + 2 \times \frac{625}{77}\right) = \frac{12200}{77} = 158\frac{34}{77} < 160,$ as required.

Comparing this with the circumference, $50\pi$, of the inscribed circle, we get the upper bound for $\pi$: $$ \pi < \frac{244}{77} = 3\frac{13}{77} < 3.17. $$

Here, the circumscribing icosagon is shown in red, with the circle (just) visible behind it in black.

irregular circumscribing icosagon

The points of tangency are marked with smaller blue circles. The 16 right-angled triangles used in construction, shown in green, are all Pythagorean, in ratio 3:4:5 or 7:24:25, with hypotenuse $\frac{625}{77}$. The other 4 edges of the icosagon are of length $\frac{50}{7}$.

There is no need to calculate any angles, but the symmetry of the figure is illuminated by knowing the sequence of angles subtended at the centre of the circle. Running counter-clockwise from the positive $x$-axis to the positive $y$-axis, it is: $\underline{\alpha\alpha\beta\beta\alpha\alpha\beta\beta\alpha\alpha}$, where $\alpha = \tan^{-1}\frac{1}{7}$, $\beta = \tan^{-1}\frac{2}{11}$, and: \begin{gather*} 2\alpha = \tan^{-1}\frac{7}{24}, \ \alpha + \beta = \tan^{-1}\frac{1}{3}, \ 2\alpha + \beta = \tan^{-1}\frac{1}{2}, \\ 2\alpha + 2\beta = \tan^{-1}\frac{3}{4}, \ 3\alpha + 2\beta = \frac{\pi}{4}, \ 4\alpha + 2\beta = \tan^{-1}\frac{4}{3}, \\ 4\alpha + 3\beta = \tan^{-1}2, \ 5\alpha + 3\beta = \tan^{-1}3, \ 4\alpha + 4\beta = \tan^{-1}\frac{24}{7}. \end{gather*}

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    $\begingroup$ Although not needed for the problem, lower bounds for $\pi$ are also easy to obtain from the figure. The inscribed icosagon gives $\pi > (6\sqrt{2}/5) + (16\sqrt{5}/25) > 3.128.$ Doubling the number of sides, using $2\sin\frac{\alpha}{2} = \sqrt{2 - 2\cos\alpha}$, etc., gives $\pi > (12\sqrt{50 - 35\sqrt{2}} + 8\sqrt{50 - 22\sqrt{5}})/5 > 3.138,$ but this is still not as good as Archimedes's amazing estimate of $\pi > 3\frac{10}{71}$ (along with $\pi < 3\frac{1}{7}$). $\endgroup$ – Calum Gilhooley Mar 16 '15 at 11:55
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Assume $\overline{AB}=1$.

Let $P$ be the intersection of the circle with $AB$.

Then $\overline{AP}\times\overline{AB}=\overline{AE}^2$ by power of point.

So $\overline{AP}=1/4$, and $\overline{PB}=3/4$. Hence the diameter of the circle is $\sqrt{(3/4)^2+1^2}=5/4$. The rest follows like the other proofs.

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    $\begingroup$ There is a typo: it should be $\overline{AO}\times\overline{AB}=\overline{AE}^2$. $\endgroup$ – user 59363 Mar 14 '15 at 8:38
  • $\begingroup$ @user59363 Yes, thanks for spotting. And I renamed $O$ to $P$ :) $\endgroup$ – user1537366 Mar 14 '15 at 13:50
  • $\begingroup$ Belated +1. So short and sweet, it almost disappeared in the deluge of proofs. Deserves more upvotes. $\endgroup$ – Calum Gilhooley Mar 17 '15 at 7:31
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enter image description here

A pure geometric solution (I think it's the same as @baum's solution and others but with details and a clarifying drawing)

Let's consider that the length of the square is 1. We can easily deduct that the square perimeter is 4.
Let (HI) be a line perpendicular to (AD), with I being its intersection with the circle).
Since (AD) is tangent to the circle, [HI] is its diameter.
$ J\hat H B =90°-J\hat B H$ (angles of a right triangle)
$J\hat B I =90°-J\hat B H$ (they both form the angle $I\hat B H=90°$ because IBH is a right triangle (inscribed in a semicircle))
therefore $ J\hat H B = J\hat B I = \theta$
$\tan(\theta)=\frac{IJ}{BJ} =2\times x$
$\tan(\theta)=\frac{BJ}{HJ} =\frac{1\over 2}{1}=\frac12$ (BJ=AH=$\frac{AD}{2}$=$\frac12$ since H is the midpoint of [AD])
$2\times x=\frac12\Rightarrow x=\frac14$
d=HI=1+$\frac14$=$\frac54$
perimeter$=2\pi r=d\pi=\frac{5\pi}{4}$

Let's compare $\frac{5\pi}{4}$ and 4
$\frac{5\pi}{4}-4=\frac{5\pi-16}{4}=\frac54(\pi-\frac{16}{5})=\frac54(\pi-3.2)<0 \text{ because } 3.2>\pi$
So $\frac{5\pi}{4}$<4
Therefore the perimeter of the square is bigger

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Put the four corners at $(0,\pm1)$ and $(2,\pm1)$. Let the center of the circle be $(x,0)$. Then the radius of the circle is $x$ and that must be the distance from the center $(x,0)$ to the corner $(2,1)$. But that latter distance is $$ \sqrt{(2-x)^2 + 1^2}. $$ Therefore \begin{align} x & = \sqrt{(2-x)^2 + 1^2} \\ x^2 & = (2-x)^2+ 1 \\ x^2 & = 4-4x+x^2 + 1 \\ 0 & = 5-4x \\ x & = 5/4. \end{align} That's the radius, so we have $$ \text{circumference}=\frac 5 4 \cdot 2\pi \approx 7.85. $$ The perimeter of the square is $8$, so that's a bit bigger.

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Using the Pythagorean theorem and the Quadratic Equation

  1. Use a unit circle. Radius of circle = 1, perimeter = 2𝜋 ~ 6.283.
  2. Center of circle = (0, 0). Point E = (-1, 0). Point B = (x, y).
  3. Perimeter of square = 4(1+x)=8y.
  4. 1 + x = 2y.
  5. 1 + 2x + x² = 4y².
  6. x² + y² = 1.
  7. 4y² = 4 - 4x².
  8. 1 + 2x + x² = 4 - 4x².
  9. 5x² + 2x - 3 = 0.
  10. x = (-2 ± √(4+60)) / 10 = (-2 ± 8) / 10 = (-2 + 8) / 10 = 0.6
    (We can rule out x = (-2 - 8) / 10 = -1, which is Point E.)
  11. y = 0.8
  12. 1 + x = 1.6 = 2 * y (also a check-by-substitution)
  13. Perimeter of square = 6.4 > Perimeter of circle ~ 6.283
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Unless you consider the Pythagorean theorem to be "a complicated geometric formula":

$$ s = \text{side length of square}\\ r = \text{radius of circle}\\ O = \text{center of circle}\\ G = \text{midpoint of BC}\\ $$

$$ r^2 = OG^2 + BG^2\\ r^2 = OG^2 + \left(\frac{s}{2}\right)^2\\ r^2 = (s - r)^2 + \frac{s^2}{4}\\ r^2 = s^2 - 2rs + r^2 + \frac{s^2}{4}\\ 0 = s^2 - 2rs + \frac{s^2}{4}\\ 2rs = \frac{5}{4} s^2\\ $$ Since $s > 0$:

$$ 2r = \frac{5}{4}s\\ r = \frac{5}{8}s\\ $$

Therefore the circumference of the circle is $2πr$; the perimeter of the square is $4s$.

$$ 2πr\stackrel{?}=4s\\ 2π \cdot \frac{5}{8}s \stackrel{?}= 4s\\ \frac{5π}{4} < 4 $$

Therefore, the circumference of the circle is less than the perimeter of the square.

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My apologies if this is already among the answers; I looked but didn't see it. (Update: Calum Gilhooley kindly informs me that it does appear among the answers, as a comment by Tebbe which simplifies Steven Stadnicki's answer.)

For convenience, let the sides of the square have length $2$. The horizontal midline of the square extends as a diameter of the circle:

enter image description here

By the intersecting chords theorem, we have

$$2(D-2)=1\cdot1$$

so $D=5/2$, and the inequality $\pi\lt3.2=16/5$ is enough to show

$$\pi D\lt 8$$

where $8$ is the perimeter of the square.

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    $\begingroup$ This is indeed nice, but I'm afraid it was already posted by Tebbe, as a very brief comment on Steven Stadnicki's solution. You are excused for missing this - because so did I, and I posted a much worse solution than yours or his! :) $\endgroup$ – Calum Gilhooley Mar 19 '15 at 20:48
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    $\begingroup$ @CalumGilhooley, ah, thank you. It was hidden from view under the "show more comments" option. BTW, I admire your effort to prove the needed inequality on $\pi$. $\endgroup$ – Barry Cipra Mar 19 '15 at 20:59
  • $\begingroup$ @CalumGilhooley, if one more person clicks on the "up" arrow beside Tebbe's comment, it should appear "above the fold" so it won't be hidden from view. $\endgroup$ – Barry Cipra Mar 19 '15 at 21:59
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This is also "computational" and not purely geometic. Let the side of the square be $1$. Let $O$ be the centre of the circle and $G$ be the middle of the segement $\overline{CB}$; then $r=\vert OB\vert$ is the radius of the circle and $1-r= \vert OG\vert $. By Pythagoras we have $(1-r)^2+(\frac{1}{2})^2=\vert OG\vert^2+\vert GB\vert^2=\vert OB\vert^2=r^2$ from which one gets $r=\frac{5}{8}$. Again one gets the same as in the answer by abel. PS I also would like to see a purely geometric argument.

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  • $\begingroup$ The almost self-effacing comment doesn't do this justice. I wonder if some notion of Greek "geometric algebra" could render such "computational" proofs (including my own unnecessary addition to the thread - in the same vein, but overcomplicated) into a less "computational", more "geometric" form? But that's a controversial and underdeveloped area, and too much to explore in answers to specific problems, or in comments like this. $\endgroup$ – Calum Gilhooley Mar 17 '15 at 8:31
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In the comments @Sid asked me to write this method down as an answer, even though it uses a formula.

If the students studied some Euclidean geometry, probably they know that the circumradius of a triangle with sides $a,b,c$ and area $S$ is $R=\frac{abc}{4S}$.

The circle drawn in the picture is the circumcircle of $BCE$. Let us choose units so that the square has side $2$ (to avoid denominators). Then $BE=CE=\sqrt{5}$, $BC=2$, $S_{BCE}=2$ are easy to find. Plugging them in the formula above, we get $R=\frac{5}{4}$.

So we have to compare $2\pi R = \frac{5}{2}\pi$ and $8$.

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    $\begingroup$ I feel like this is the real answer high school kids would come up with, and the one they'd be expected to use. I think the key to OP's question is that he wanted to omit complicated formulas. This one isn't complicated, it just requires that the students have covered circumcircles. Easily solvable under a minute with the right tool. $\endgroup$ – Shaz Mar 19 '15 at 20:11
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This calculation uses only the upper limit on the value of pi calculated by Archimedes (22/7) and simplifies the algebra somewhat by using an intermediate distance value, that of the base ZM of the triangle ZMB.


Define points Z as the centre of the circle and M as the midpoint of the line segment BC.

Now applying Pythagoras to the triangle ZMB in terms of a (the length of the segment ZM) and r (radius of the circle) yields (since L, the side length of the square, $ = r + a$):
$$ r^2 - a^2 = (r + a)^2 / 4 $$ and dividing by (r + a) gives $$ r - a = (r + a) /4 $$ which upon rearrangement and substitution of L back in gives $$ 3r = 5a = 5 (L - r) $$ and finally $$ 8 r = 5 L $$ Then the ratio of the circumference of the circle to that of the square is obviously $$ ratio = 2 \pi r / (4 *(8/5) r) = 10 \pi / 32 = 5 \pi / 16 $$ However, since $\pi < 22/7$ (by Archimedes), the upper limit of this ratio is $$ ratio = 5 \pi / 16 < 5 * 22 / (16 * 7) = 110/112 \approx 0.98 $$

Therefore the square has the greater circumference.

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    $\begingroup$ I might expect a high school student to prove division by $(r+a)$ is valid (neither are $\leq 0$). And $perimeter of square = 32/5 r = 6.4 r$ explicitly might make people happy. The ratio math to show that $6.4 > 2 \pi$ while cute might even be omitted at that point. $\endgroup$ – Yakk Mar 17 '15 at 13:39
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You can solve this using just the quadratic formula and the equation of a circle, without bringing triangles, Pythagoras, or real geometric reasoning into it at all.

These are common enough hammers for all problems in high school, where everything turns out to be quadratic equations in disguise. Everything falls out nicely enough that it could even be what's intended.

  1. Assume a unit circle centred at the origin. $P_{\mathrm{circle}} = 2\pi$.
  2. $E$ is at $(-1, 0)$.
  3. $A$ is at $(-1, h)$.
  4. Note that $h$ is half the edge length of the square. $P_{\mathrm{square}} = 8h$.
  5. $B$ is thus at $(-1 + 2h, h)$. This expression makes the qudratic formula become trivial.
  6. Substitute these coordinates into $x^2 + y^2 = r^2$: $(-1 + 2h)^2 + h^2 = 1$.
  7. Expand and simplify: $5h^2 - 4h = 0$.
  8. So by the quadratic formula $h = {4 \pm \sqrt{(-4)^2 - 4 ~\cdot~ 5 ~\cdot~ 0} \over{2 ~\cdot~ 5}}$.
  9. The discriminant simplifies almost away to get ${4 \pm \sqrt{4^2 - 0} \over 10}$, where the square root is trivial.
  10. $h = {8\over10} \mathrm{~or~} {0\over 10}$.
  11. $h = 0.8 \rightarrow P_{\mathrm{square}} = 8h = 6.4$.
  12. $6.4 > 2\pi$.

Doing it this way also highlights the degenerate solution where the square's just a single point of mush on the perimeter of the circle, which is exactly the kind of detail I would have been expected to mention in an exam.

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    $\begingroup$ The equation of the unit circle is a special case of the Pythagorean theorem. $\endgroup$ – Jasper Mar 15 '15 at 20:02
  • $\begingroup$ The second case is where the square is point E, and the square's perimeter is zero. The circle still has its perimeter of 2π. $\endgroup$ – Jasper Mar 15 '15 at 20:07
  • $\begingroup$ @Jasper: Not generally taught as such in high school, though - they learn the equation just as a way of introducing graphs in x and y (at least I did). The other point, yes, of course, that was silly. $\endgroup$ – Michael Homer Mar 15 '15 at 20:17
  • $\begingroup$ The opening sentence is somewhat misleading. You must use the quadratic equation and an reasonable approximation (more precisely upper bound) for the value of$~\pi$. $\endgroup$ – Marc van Leeuwen Mar 16 '15 at 13:33
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Using standard exam guesses

This happens to boil down to a 3:4:5 right triangle, so guessing can be used to solve the problem.

  1. Use standard unit circle.
  2. Perimeter of unit circle = 2𝜋 ~ 6.283.
  3. Use standard unit circle (cartesian) coordinate system. Point E is at (-1, 0).
  4. Let point O be the center of the circle, at (0, 0).
  5. Let point F be the midpoint of the right side of the square.
  6. Consider a few possible shapes for the right triangle OFB:

45-45-90 right triangle
Does not work, because a square based on this triangle does not reach to the left edge of the unit circle.

30-60-90 right triangle
Height BC is √3.
Width AB=EF is 1.5.
Does not work, because it is not a square.

3:4:5 right triangle
Height BC is 2 * 0.8 = 1.6
Width AB=EF is 1 + 0.6 = 1.6
Works; this is a square. Square perimeter = 6.4 > Circle perimeter = 2𝜋 ~ 6.283.

5:12:13 right triangle Don't need to evaluate, because the 3:4:5 right triangle solved the problem.

8:15:17 right triangle Don't need to evaluate, because the 3:4:5 right triangle solved the problem.

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You can show this with a ruler-and compass construction, in the spirit of Archimedes. Or, you can get the same result with simple trig and a calculator, measuring angles in degrees only, without using calculus or knowing the value of $\pi$.

As the other answers have shown, we are comparing the perimeters of a square of side 8 and a circle radius 5.

If we circumscribe a regular polygon with $n$ sides around a circle radius $r$ so that the circle touches the mid-point of each side, each side of the polygon is length $2r \tan (180/n)^\circ$ and the circumference of the polygon is $2nr \tan (180/n)^\circ$.

It is obvious (at high school level!) that the circumference of the polygon is longer than the circumference of the circle.

So, start by circumscribing a square around the circle, and keep doubling the number of sides till something interesting happens.

$r = 5$.

Square: $n = 4$, perimeter = $2 \times 5 \times 4 \tan 45^\circ = 40$, which is a sanity check when using the calculator, of course.

Octagon: $n = 8$, perimeter = $2 \times 5 \times 8 \tan 22.5^\circ = 33.1$.

16-gon: $n = 16$, perimeter = $160 \tan 11.25^\circ < 31.9$. So the inscribed circle perimeter is $< 31.9$, which is less than the perimeter of a square with side $8$.

For extra credit, get them to check this by making a scale drawing.

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The problem is simply solved if the relative position of the center and the radius of the circle are known. The attached shows how they can be obtained.

enter image description here

Added:

Let the length of a side of the square be $L$. Then, the radius of the circle $= (\dfrac {5}{4}L) \times (\dfrac {1}{2})$.

$\dfrac {perimeter_{circle}}{perimeter_{square}} = ... = \dfrac {\pi}{3.2} \lt 1$.

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    $\begingroup$ This is good, but how do you know that AB' is exactly one quarter of AB? $\endgroup$ – curiousdannii Mar 15 '15 at 23:26
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    $\begingroup$ @curiousdannii By symmetry. Notice that (1) this question was given to high school student; and (2) “this problem is supposed to be very simple; it shouldn't require the student to know the formula …… to solve the problem without knowing any complicated geometric formulas”. $\endgroup$ – Mick Mar 16 '15 at 5:47
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    $\begingroup$ I've made use of your diagram in a comment on my own answer, proving that $AB' = A'B$ and $DC' = D'C$ - which I'd laboured to do before. Thank you for that! However, I don't understand how your diagram proves that $AB'$ is a quarter of $AB$. My argument also seemed to need a further step, in which I used Pythagoras to prove that the ratio is 4; was this step not necessary? $\endgroup$ – Calum Gilhooley Mar 16 '15 at 21:32
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    $\begingroup$ @CalumGilhooley You are welcome to do so. There are many ways to prove AB’ = AB/4 but most of them are tedious. The shortest (but not the simplest) way to prove that is by “power of a point” which, unfortunately, could be beyond the OP’s intention. $\endgroup$ – Mick Mar 17 '15 at 4:17
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    $\begingroup$ @CalumGilhooley To prove AB' = BA', we can do the following:- Let M and M’ be the midpoints of BC and B’C’ respectively. Then, AB’ = EM’ = ME’ = BA’. The middle equality comes from “Radius – distance from center to the midpoints of equal chords”. $\endgroup$ – Mick Mar 17 '15 at 4:42
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Another late entry (I just saw this question in the Hot Topics list), but I arrived at this solution before looking at the other answers. It roughly follows Sid's suggestion in the question; I don't think it's beyond the abilities of the high school students that the problem is aimed at.

Let the square's side length be 2, with the square centered on the origin, so the coordinates of the corners are (±1, ±1). B is (1, 1) and E is (-1, 0).

By symmetry, the centre of the circle M is on the X axis; let M be ($x$, 0).

Let $r$ be the radius of the circle.

$r = |ME| = x - -1 = x + 1$

By the Pythagorean distance formula:

$r^2 = |MB|^2 = (x - 1)^2 + (0 - 1)^2 = (x - 1)^2 + 1$

Thus $$\begin{align} (x + 1)^2 & = (x - 1)^2 + 1 \\ x^2 + 2x + 1 & = x^2 - 2x + 1 + 1 \\ 4x & = 1 \\ x & = \frac{1}{4} \\ r = x + 1 & = \frac{5}{4} \\ \end{align}$$

So the circumference $2\pi r = \frac{5\pi}{2}$. Now $\pi < \frac{22}{7}$ so the circumference is less than $\frac{5 \times 22}{2 \times 7} = \frac{55}{7}$, which is less than 8, the perimeter of the square.

No quadratic or obscure geometrical formulae required, nor the ability to magically see 3, 4, 5 triangles. :)

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Redrawn diagram above. OP picture

The $\measuredangle$CDF is determined and is given by: $$\arctan(\frac{a}{a/2}) = 63.4 ^ \circ$$ Then the $\measuredangle$FDB can be determined as:

$(180 - 2\times63.4) ^ \circ = 53.2 ^ \circ$, because $\measuredangle$CDF is equal to $\measuredangle$EDB, and the three angles involved add up to $180^ \circ$.

We then have:

$\measuredangle$FAB is 2 times that of $\measuredangle$FDB by the incribed angle theorem.
$\measuredangle$FAG is 0.5 times that of $\measuredangle$FAB.

Hence, $\measuredangle$$FAG$ $=$ $\measuredangle$$FDB$ $=53.2 ^ \circ$.

Finally, in the right angled triangle FAG, we obtain the relationship between the radius $r$ and the side a of the square as: $$\sin\space(53.2 \space^ \circ) = \frac{a/2}{r}$$ From which it follows that: $1.25 \times a=2r$. Hence, the perimeter of the circle is $\pi 2r = 3.92 a$, which is less than $4a$, the perimeter of the square.

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A little late to the party, but a slight variation:

I drew the same diagram as Steven Stadnicki, but used it differently to find the diameter of the circle:

enter image description here

Define $s = |AB|$, the length of a side of the square.

Draw the diameter from $E$, intersecting the square at $G$ and the circle at $F$. Clearly $|EG|=s$ and $|BG| = s/2$.

Consider the triangles $\triangle EBG$ and $\triangle EBF$. These are both right angle triangles with the common angle at $E$, so are similar. Therefore $\dfrac{|BE|}{|EG|} = \dfrac{|EF|}{|BE|}$, so $ |EF| = \dfrac {|BE|^2}{s}$.

By Pythagorus, $|BE|^2 = s^2 + (\frac{1}{2}s)^2 = \frac{5}{4}s^2$. So $|EF| = \frac{5}{4}s$.

Then the perimeter of the square, $4s = \frac{16}{5}\frac{5}{4}s = 3.2\frac{5}{4}s$, is greater than $\pi\frac{5}{4}s$, the perimeter of the circle.

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Here's another proof:

  • Denote by $a$ the side of the square

  • Take the point $F$, like in the figure. Since $\angle CBF = 90^\circ$, it follows that $CF$ is a diamenter of the circle.

  • Then $\angle CEF = 90^\circ$ and triangles $\Delta DCE$ and $\Delta AEF$ are similar. This implies that $AF = a/4$ and thus $BF = 3a/4$,

  • Now, apply Pythagora's theorem to find $CF = 5a/4$. This gives the radius of the circle, and the answer can be found immediately.

enter image description here

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Let the circle be the unit circle. Consider the square of equal perimeter: it has side length $\pi/2$. Align this square so that the midpoint of its left side touches the leftmost point of the unit circle. Where is the upper right corner of the square? It has location $(-1+\pi/2,\pi/4)$, which is at distance $\sqrt{1-\pi+5\pi^2/16}<1$ (since $\sqrt{1-\pi+5\pi^2/16}=\sqrt{1-\pi(1-5\pi/16)}$, and $5\pi/16<5*3.2/16=1 \implies 1-5\pi/16>0 \implies 1-\pi(1-5\pi/16)<1$) from the center of the circle. Hence you must make this square bigger to get it to touch the circle. Thus the square of the original problem is bigger, and so it has bigger perimeter than the circle.

enter image description here

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Here's another co-ordinates-free approach:

In the OP's diagram, let the centre of the circle be $O$ then clearly triangle $AOE$ is congruent to triangle $OED$ and triangle $ABO$ is congruent to triangle $CDO.$ Further, $AB, OE$ and $DC$ are all parallel. Together this implies that the area of trapezium $ABOE$ is equal to that of trapezium $CDEO.$ Clearly the area of both the trapeziums plus the isosceles triangle $BOC$ is equal to the area of the square. If the radius of the circle is $r$ so that $EO=OB=OC=r$ we then have:

$\Rightarrow \text{Ar}[ABOE] + \text{Ar}[CDEO] + \text{Ar}[BOC] = \text{Ar}[ABCD]$

$\Rightarrow 2(r+AD)\dfrac{AD}{2} + \dfrac{AD}{2}\left(\sqrt{r^2-\dfrac{AD^2}{4}}\right)= AD^2$

$\Rightarrow r =\dfrac{5AD}{8}$

$\Rightarrow \dfrac{\text{perimeter of square}}{\text{circumference of circle}} =\dfrac{4AD}{\dfrac{5\pi AD}{4}} = \dfrac{16}{5\pi}>1$

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protected by Community Mar 17 '15 at 15:22

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