10
$\begingroup$

Two side-by-side squares are inscribed in a semicircle. The diameter of the semicircle is $16$. What is the sum of the two squares' areas?

This is a bonus question from my online math class that I've been trying to solve but haven't gotten very far. I named $CD$ $x$ and $HD$ $y$, then drew $OB$ and $OF$, then did the Pythagorean theorem to try to get $x^2+y^2$ but I didn't have much luck with that.

Diagram

$\endgroup$
  • $\begingroup$ I think it probably has to do with this $\endgroup$ – saulspatz Sep 22 at 16:35
  • 8
    $\begingroup$ Take the specific case where the squares are congruent. The solution in that case is trivial. $\endgroup$ – Daniel Mathias Sep 22 at 16:52
  • $\begingroup$ @DanielMathias But then you would still need to show that the sum is independent of $D$. $\endgroup$ – Teepeemm Sep 23 at 3:00
  • $\begingroup$ @Teepeemm This is an online thing, it's probably just a field where you type in the answer. $\endgroup$ – MattPutnam Sep 23 at 3:38
  • $\begingroup$ All the solutions given (and indeed the question, I suppose) assume that length OD <= length OA. If the right hand square is inscribed within the semi circle and becomes bounded by C0 constrained to r, the solution is not independent of D. $\endgroup$ – Phil H Sep 23 at 12:21
7
$\begingroup$

Let $OD=c$ and the side lengths of the two squares $a$ and $b$. From right triangles OAB and OHF,

$$r^2=a^2+(a-c)^2;\>\>\>\>\> r^2=b^2+(b+c)^2$$

Eliminate $c$ to get,

$$a-\sqrt{r^2-a^2}=\sqrt{r^2-b^2}-b$$

Square both sides,

$$a\sqrt{ r^2-a^2} =b\sqrt{r^2-b^2}$$

Square again and rearrange,

$$r^2(a^2-b^2)=a^4-b^4$$

Thus, the sum of the two areas is

$$a^2+b^2=r^2=64$$

$\endgroup$
8
$\begingroup$

A somewhat devious way is to extract the fact that the answer doesn't depend on where D is, so place it at the origin!

Then it's simply:

$$ x = y $$

$$ \therefore x^2 + y^2 = 2 x^2 = 8^2 $$

$\endgroup$
  • $\begingroup$ That is perhaps the neatest use of W.L.O.G I've seen, definitely using that as an example for why we can use it in future. $\endgroup$ – Fifth_H0r5eman Sep 23 at 12:07
  • $\begingroup$ The "without loss of generality" is implicit in the phrasing of this question. A more devious question would be to ask what the maximum area is (Still 64, but now you've got to prove it takes that value for |OD|<=|OA| ) $\endgroup$ – MSalters Sep 23 at 12:26
  • $\begingroup$ @Fifth_H0r5eman This is a pretty terrible example of WLOG because it's not clear that the area is independent of D. Assuming it is works as a technique for contest math because you know that you've been given enough information to solve the problem, but in general making random assumptions like this will just lead to mistakes. $\endgroup$ – Brady Gilg Sep 23 at 17:04
  • $\begingroup$ Given that no lengths are given obviously D is arbitrarily placed, so assuming AD = HD is completely fine. $\endgroup$ – Fifth_H0r5eman Sep 23 at 17:47
  • $\begingroup$ @Fifth_H0r5eman It's obvious that D is arbitrarily placed. Is it obvious that the placement of D doesn't change the area? $\endgroup$ – Brady Gilg Sep 23 at 23:06
6
$\begingroup$

enter image description here

Let $V$ be the common vertex of our squares which lies on the diameter of the semicircle. We may assume that the coordinates of $V$ are $(x,0)$. The upper-right-corner $A$ is located at the intersection (the one with a positive ordinate) of $x^2+y^2=64$ and $y=x-v$. The upper-left-corner $B$ is located at the intersection (always the one with a positive ordinate) of $y=v-x$ and $x^2+y^2=64$. It follows that the ordinate of $A$ is $\frac{1}{2}\left(-v+\sqrt{128-v^2}\right)$ and the ordinate of $B$ is $\frac{1}{2}\left(v+\sqrt{128-v^2}\right)$. By summing the squares of these numbers we get that the total area of our squares is $$ \frac{1}{4}(v^2+128-v^2+v^2+128-v^2) = 64,$$ i.e. the area of a square built on a radius.

enter image description here

In order to produce an elementary proof, we just have to show that the length of $AB=\sqrt{2}\sqrt{AA'^2+BB'^2}$ does not depend on the position of $V$ on the diameter. But this is trivial since the symmetric of $B$ with respect to the diameter "sees" the $AB$-chord under an angle of $45^\circ$, such that $\widehat{AOB}$ always is a right angle (and $AOVB$ is a cyclic quadrilateral).

$\endgroup$
5
$\begingroup$

You can use the Pythagorean theorem twice: enter image description here

$\endgroup$
  • 9
    $\begingroup$ It needs to be mentioned why OB is perpendicular to OF. $\endgroup$ – Narasimham Sep 23 at 6:19
1
$\begingroup$

The base $CA$ is partitioned into $(a,b)$ segments for the common squares corner and $(b,a) $ for locating center of circle.

To appreciate this direct analytical geometry is enough:

Equation of perpendicular bisector from the sketch:

$$\dfrac{ y-(a+b)/2}{x-(a+b)/2}= \dfrac{ (a+b)}{(a-b)}$$

At y=o, x=? Solving

enter image description here

$$ x_O=b $$

EDIT1:

So in the construction all you need to find the center of circle is marking off segment length $b$ with a compass centered at $C$ and complete the circle through square outer vertices $(F,B)$ of required circle radius $R$ centered at $O$.

Shown case is chosen $ (a=3,b=4)$ from familiar Pythagorean triplet with diameter of circle $10$ instead of $16$

$$ R^2= a^2+b^2 = R^2= 5^2 $$

The squares area sum equals area of square made on a side of radius $R$

$\endgroup$
  • $\begingroup$ Upvoting because this construction of the line equidistant from two points, and its point of intersection with the baseline that the two squares share, is really key for my understanding the possibilities here. You have three points B,C,F and this construction potentially creates three circles which can inscribe the two squares: one from this construction applied to {B, C}, one from it applied to {C, F}, one from it applied to {B, F}. It seems plausible that at any time only two of these lead to distinct inscriptions, but only the {B,F} one has the given property, if it is even valid. $\endgroup$ – CR Drost Sep 24 at 14:33
  • $\begingroup$ Thanks for comment. $C,A$ are ruled out from given problem geometry. I added edit about how to locate $0$ $\endgroup$ – Narasimham Sep 24 at 20:18
0
$\begingroup$
  1. We know the line between the semi circle and D exactly bisects each square, and thus FDB forms a right angle.
  2. We know that for some angle FDM, the y coordinate at F is $sin \theta$
  3. The right angle FDB then requires that the y coordinate for B is $sin (\theta + \pi/2) = cos \theta$
  4. The area of each square is the square of those y coordinates, and thus the sum is $(r sin \theta)^2 + (r cos \theta)^2$
  5. Given the identity $sin^2 \theta + cos^2 \theta = 1$, we can simplify the result to $r^2 = 64$.

Incidentally if you combine this approach for the unit circle with the Pythagorean approach in other answers, you have a neat proof of the trig identity we used. It also demonstrates that all such paths between points on a circle which form a right angle have equal length.

A simpler form is to note #1 as above, then:

  1. All valid combinations of F and B form a right angle triangle with the same length hypotenuse.
  2. The square of that hypotenuse is always the same as the required area by substituting for the squares' diagonals, and thus the result is constant.
  3. If the result is constant it must equal the easy symmetric case and thus equals $r^2$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.