Here is the picture of the question:

geometry question

  • $ABC$ is a right triangle.
  • $m(CBA)=90^\circ$.
  • $m(BAD)=2m(DAC)=2\alpha$.
  • $D$ is a midpoint of $[BC]$.
  • $E$ is a point on $[AD]$.
  • $m(BED)=90^\circ$.
  • $|DE|=3$.
  • What is $|AB|=x$?

Tried lots of things which gives me some trigonometric identities, but none of them led me to the solution which is $x=6$. And even if it did, i prefer more geometric methods (all solutions are welcome though). Note that this could be an easy question, and i could probably missing something obvious.

  • $\begingroup$ Thoughts: use law of sines to relate $\alpha$ and $y$ (the length of $DC$) to $3\alpha$ and $\sqrt{x^2+y^2}$; similar triangles to relate $x$ and $y$ to $3$ and $\sqrt{y^2-9}$; and definition of $\cos 2\alpha = 3/y$. $\endgroup$ – Greg Martin Jan 25 '16 at 21:26
  • $\begingroup$ @GregMartin, already did that. I even obtained all segments in terms of $\sin\alpha$ hoping i will reach something nice. There is a nice formula though, attainable by the law of sines to left and right triangle, which is $\cos\alpha\cos3\alpha=\frac{1}{2}$. $\endgroup$ – Alistair Jan 25 '16 at 21:42
  • $\begingroup$ What is the origin of this question? $\endgroup$ – Brian Tung Jan 25 '16 at 23:19
  • $\begingroup$ @BrianTung, most likely, came from some test for preparing high school students to university exams. This seems like a special triangle so i searched google a little before asking, but nothing comes up. I also found $\cos2\alpha=\frac{\sqrt{17}-1}{4}$ which strengthens my view that this is a special triangle that i don't heard about. $\endgroup$ – Alistair Jan 26 '16 at 1:18
  • $\begingroup$ But where did you find it? $\endgroup$ – Brian Tung Jan 26 '16 at 1:22

enter image description here

(1) The brown dotted line is the extension of $BE$.

(2) The green dotted line ($AC’$) is the angle bisector of $∠BAE$ such that $\beta_1 = \beta_2 = \beta$. Another obvious fact is $\beta ‘ = 2\beta$.

(3) The blue dotted line is the perpendicular bisector of $AB$ cutting $AB$ and $AC$ at $P$ and $Q$ respectively.

By intercept theorem, $AQ = QC$.

By midpoint theorem, $QD = BP = PA$.

$L$ is point on $BC$ such that $EL \bot BC$. $EL$ is extended to cut $AC’$ at $N$. It should also be clear that $\angle 1 = \angle 2 = \beta’$..

The line $AEDM$ is the axis of symmetry for $\triangle ACC’$. Then, $\angle AMC = 90^\circ$. This means M is also on the circum-circle (centered at Q) of $\triangle ABC$.

Therefore, $\triangle BED \equiv \triangle CMD \Rightarrow DE = DM$.

$QA = QM \Rightarrow \beta_3 = \beta$. $\beta_4 = \angle 2 - \beta_3 = \beta \Rightarrow QD = DM$.

Result follows.

  • $\begingroup$ That's the geometric approach i'm looking for. I need clarification on one part though: How do you reach that $M$ is on the $Q$ centered circumcircle from $m(AMC) = 90^\circ$? $\endgroup$ – Alistair Jan 28 '16 at 6:28
  • $\begingroup$ @Alistair You can prove that triangles $AE(?)$ and $AE(??)$ are congruent. From that, you can further deduce triangles $AMC$ and $AMC’$ are congruent [Note that AC = AC’ because C’ is the intersection of the angle bisector of $\angle BAD$ and the circle (centered at A and radius = AC). Hence $\angle AMC = 90^\circ$. Or simply just by noting that AM is the axis of symmetry for the $\triangle ACC’$. $\endgroup$ – Mick Jan 28 '16 at 7:49
  • 1
    $\begingroup$ @Alistair Then, $\angle M = \angle B = 90^0$ implies M is also on that circle. This is the converse of "angles in the same segment". $\endgroup$ – Mick Jan 28 '16 at 7:56
  • $\begingroup$ Ah, i see. I failed to see $\angle M$ and $\angle B$ see the diameter of $Q$ centered circle. But i believe your answer may be shortened. It seems you flip the $\triangle BDE$ to create $\triangle CDM$, then observe $M$ is on circle (angles with equal measures looking same arch). There is no need to angle bisector or axis of symmetry. $\endgroup$ – Alistair Jan 28 '16 at 16:12
  • $\begingroup$ @Alistair Flipping is not a part in the old Euclidean geometry. The deduction yielded is then not that convincing, especially when it is not backed up by known theorems as reasons. Also, it is almost clear that the angle bisector is necessary when one angle is the double of the other. $\endgroup$ – Mick Jan 28 '16 at 16:58

This is one approach. I don't pretend it's the nicest.

Let $u = \tan\alpha$. Then

$$ \tan2\alpha = \frac{2u}{1-u^2} $$ $$ \tan3\alpha = \frac{u(3-u^2)}{1-3u^2} $$

Since $\tan3\alpha = 2\tan2\alpha$, we have

$$ \frac{4u}{1-u^2} = \frac{u(3-u^2)}{1-3u^2} $$ $$ (1-u^2)(3-u^2) = 4(1-3u^2) $$ $$ 3-4u^2+u^4 = 4-12u^2 $$ $$ 1-u^4 = 8u^2 $$

Now let $y = BE$. By similar triangles, we have

$$ \frac{3}{y} = \frac{\sqrt{9+y^2}}{x} = \tan2\alpha = \frac{2u}{1-u^2} $$

Square all terms to obtain

$$ \frac{9}{y^2} = \frac{9+y^2}{x^2} = \left(\frac{2u}{1-u^2}\right)^2 = \frac{4u^2}{1-2u^2+u^4} $$

Observe that the outer pair of the equality above gives us

$$ \frac{9+y^2}{y^2} = \frac{9}{y^2}+1 = 1+\frac{4u^2}{1-2u^2+u^4} = \frac{1+2u^2+u^4}{1-2u^2+u^4} = \left(\frac{1+u^2}{1-u^2}\right)^2 $$

Multiplying both ends by $\frac{x^2}{9+y^2} \cdot \frac{y^2}{9} = \left(\frac{1-u^2}{2u}\right)^2\left(\frac{1-u^2}{2u}\right)^2$ yields

\begin{align} \frac{x^2}{9} & = \left(\frac{1+u^2}{2u}\right)^2 \left(\frac{1-u^2}{2u}\right)^2 \\ & = \left(\frac{1-u^4}{4u^2}\right)^2 \\ & = \left(\frac{8u^2}{4u^2}\right)^2 \qquad \longleftarrow 1-u^4 = 8u^2 \\ & = 2^2 = 4 \end{align}

So $x^2 = 36$ and $x = 6$.

Neat question. One almost feels as though the simple and straightforward answer must be obtainable with a correspondingly simple and straightforward approach. Thus far, though, I haven't seen it. Anybody?

  • $\begingroup$ Nice manipulations. I can see why you use $\tan\alpha$ instead of $\sin\alpha$ and $\cos\alpha$ like i did. Though i am doubtful if i can make these manipulations and reach the answer even if i used $\tan\alpha$. I like this question too and i still believe there is a geometric answer to it. $\endgroup$ – Alistair Jan 25 '16 at 23:18
  • $\begingroup$ I'll keep an eye out here to see if anyone comes up with anything. $\endgroup$ – Brian Tung Jan 25 '16 at 23:24

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