# A concise distance problem

A falsely simple Euclidian geometry problem:

Points $A$, $B$, $C$ are collinear; $\|AB\|=\|BD\|=\|CD\|=1$; $\|AC\|=\|AD\|$.
What is the set of possible $\|AC\|$ ?

I'm after a concise answer, with reasoning, that would get maximum points to an 11th-grader.

A related question asks an appropriate level for the problem (worded in less mathematical terms and thus reduced to distinct points).

To check your answer: the mean of the elements of the set of solutions to the present question is $\approx 1.08$.

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In the other question you gave two possible answers, one with $C$ on either side of $A$, which looks right to me. Here you only give one answer, and it's not one of the other two answers. The questions appear identical, though, and you seem to be implying that they're intended to be identical. Why the different answers? – joriki May 6 '13 at 6:47
You might be missing the case $B=C$ (which I believe gives a distance $1$). btw, what is a $10^{\text{th}}$ grader in your area supposed to know? Trigonometry? Analytic Geometry? Since you have termed this as a pure geometry problem, without the Alice etc, you might want to clarify that they are distinct points. – Aryabhata May 6 '13 at 7:11
@Aryabhata: in France, a 10-th grader in a scientific class (15-16 years, classe de seconde) knows point coordinates, computing distances, and solving a second-degree equation graphically or by successive approximation. Following that realization, I changed to 11-th grade, where I assume algebraic solution and basic trig are introduced. – fgrieu May 6 '13 at 7:39
I downvoted for deliberate obscurity. You generate the impression that the questions are the same whereas the intent is apparently that they are not, since points can coincide and persons can't. You fail to point out the differences, and the different number and character of the solutions. That turns this into a riddle instead of focusing on the actual question regarding a concise answer by a high school student. To top things off, you seem to have miscalculated; adding the third value $\lVert AC\rVert=1$ to the two you give in the other answer yields a mean of $\sqrt5/3\approx1.08$. – joriki May 6 '13 at 7:44
@fgrieu: If no obfuscation is intended, I suggest at least the following edits to the question: Point out the difference between this question and the one you link to, and clarify that the value in the spoiler box is the mean of the three possible values of $\lVert AC\rVert$. Ideally, I think there shouldn't be any spoiler boxes or means or the like at all here; you're asking for a concise answer from a high-school student and not for a solution of the problem itself, so there's no point in hiding the answer. – joriki May 6 '13 at 7:56

W.l.o.g. choose coordinates as follows:

\begin{align*} A &= \begin{pmatrix}0\\0\end{pmatrix} & B &= \begin{pmatrix}1\\0\end{pmatrix} & C &= \begin{pmatrix}x\\0\end{pmatrix} & D &= \begin{pmatrix}\frac{x+1}2\\ \sqrt{x^2-\left(\frac{x+1}{2}\right)^2}\end{pmatrix} \end{align*}

Point $D$ is choosen on the perpendicular bisector of $BC$. This bisector is only uniquely defined and a strict requirement if $B\neq C$, so the case of $B=C$ will have to be handled separately. It is furthermore chosen at distance $x$ from $A$. So the above coordinates will already ensure the following conditions:

\begin{align*} \lVert AB\rVert &= 1 & \lVert BD\rVert &= \lVert CD\rVert & \lVert AC\rVert &= \lVert AD\rVert \end{align*}

Now all that remains is the condition $\lVert BD\rVert = 1$ or equivalently $\lVert CD\rVert = 1$. I go for the latter.

\begin{align*} 1 = \lVert CD\rVert^2 &= \left(x-\frac{x+1}2\right)^2 + \left(\sqrt{x^2-\left(\frac{x+1}{2}\right)^2}\right)^2 \\&= \left(\frac{x-1}2\right)^2 + \left(x^2-\left(\frac{x+1}{2}\right)^2\right) \\ 4 &= \left(x^2-2x+1\right) + \left(4x^2-x^2-2x-1\right) \\ 0 &= 4x^2 - 4x - 4 = x^2 - x - 1 \\ x_{1,2} &= \frac{1\pm\sqrt{1+4}}{2} \\ x_1 &= \frac{1-\sqrt5}2 \approx -0.618 \\ x_2 &= \frac{1+\sqrt5}2 \approx 1.618 \end{align*}

The special case of $B=C$ gives a third solution $x_3=1$ describing $C$, with an associated point $D_3=\begin{pmatrix}\frac12\\\frac{\sqrt3}2\end{pmatrix}$ (although these coordinates of $D_3$ are not neccessary to answer the question). The requested length is the absolute value of $x$, so the final solution is

$$\lVert AC\rVert = \lvert x\rvert \in \left\{ \frac{\sqrt5-1}2, 1, \frac{\sqrt5+1}2 \right\}$$

The idea of choosing suitable coordinates without loss of generality will probably be the most problematic concept. Whether students will be able to think of this on their own, particularly in an exam situation, depends a lot on the details of their education. After that, chances are good that many will miss the special case, since the coordinates (as perhaps derived from a sketch of a non-degenerate situation) seem to perclude such a third solution. The computation of $x$ on the other hand should be pretty straight forward, once they got this far.

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Nice. The algebra may be a tad simpler with the origin at $B$. – fgrieu May 6 '13 at 14:20

If $D$ was on the line $AB$, then either $D = A$, $B = C$, or $A, B, D, C$ would on a line (in the order). $D = A$ is impossible because that would imply $C = A$ and $||CD|| = 0$, and $B = C$ is impossible because that would imply $||AD|| = 2$but $||AC|| = 1$. The last case is impossible, because $||AC|| \neq ||AD||$.

Therefore, $D$ is not colinear with $A, B, C$. Additionally, $C \neq A$.

(edit: When $C = A$, $||CB|| = ||BD|| = ||DC|| = 1$, so an ᅟequilateral triangle $\triangle CBD$ can be formed.)

If $C$ was inside $AB$, since $D$ is on a circle with center $A$ and radius $||AC||$, projection of $D$ on $AB$ should be between $A$ and $C$. However, since $||BD|| = ||CD||$, projection of $D$ should be located at midpoint of $BC$. Therefore, it is impossible.

Therefore, either $C$ is located on [1] ray $AB$ - $AB$, or [2] ray $BA$ - $AB$.

## The first case

$\angle BAD = \angle BDA$ since $||AB|| = ||BD||$. $\angle DBC = \angle DCB$ since $||BD|| = ||CD||$. Since $\angle DBC = 2 \angle DAB$, the sum of inner angles of triangle $ADC$ can be calculated by

$$\angle DAC + \angle ADC + \angle ACD = 5 \angle DAC = 180°$$

Therefore, $\angle DAC = 36°$. Now, since $\triangle ACD \sim \triangle DBC$,

$$||AC|| : ||CD|| \equiv ||DB|| : ||BC||$$ $$||AC|| : 1 \equiv 1 : ||AC|| - 1$$

And now $||AC||$ can be calculated by solving a quadratic equation and the fact that $||AC|| > 1$.

## The second case

Do the same thing. But now $||BC|| = ||AC||+1$.

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There's a third solution with $B=C$, with $ABD$ forming an equilateral triangle with side length $1$. – joriki May 7 '13 at 12:32
@joriki Thanks. I missed that case.. – JiminP May 7 '13 at 12:51
$C=A$ does not work; $B=C$ does. But +1 nevertheless for the derivation of $\angle DAC$ – fgrieu May 8 '13 at 8:12