In a right angled $\triangle ABC$, $\angle B = 90^\circ$, $\angle C = 15^\circ$ and $|AC| = 7.\;$ Let a point $D$ (Random Point) be taken on $AC$ and then perpendicular lines $DE$ and $DF$ are drawn on $AB$ and $AC$ respectively. What is the probability of $DE\cdot DF >3?$


By trigonometry, I got the length of other two side from the hypotenuse $AC:$

$AB$ $\approx 1.8117$

$BC$ $\approx 6.7614$. And than, I got the equation that

$DE\cdot DF = (6.7614 - DE)\cdot AE\;$ (from the similarity of both $\triangle AED$ and $\triangle DFC$)

Again, from right angled $\triangle AED$,

$\dfrac{AE}{DE} = \tan 15^{\circ}\quad \implies \quad AE = DE\cdot \tan 15^\circ$

Here, I got stuck. I couldn't find a way out to proceed and skip that situation. I became lost and was unable to complete that process. Any kind of help or clue will be greatly helpful for me to step forward.

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    $\begingroup$ I say the probability is zero, because I'm going to pick point $D$ according to my own distribution, which picks point $C$ 100 percent of the time. I'm joking, of course, but the problem's incomplete: There's nothing in the problem statement that specifies the distribution from which $D$ is picked, which is essential in the problem having an actual conclusion. $\endgroup$ – John Hughes Feb 4 at 12:05
  • $\begingroup$ @Sir JohnHughes I am extremely sorry. But I chose the point $D$ according to own aspects. If the diagrams makes disturbance for proper figure or visualization, then I have nothing to do. From that specific location of $D$ (that I have picked up), we have to proceed towards the conclusion. $\endgroup$ – Anirban Niloy Feb 4 at 12:10
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    $\begingroup$ Don't be sorry -- I'm merely suggesting an improvement in the problem statement. It should be "Point $D$ is drawn from a uniform distribution on $AC$..." or something like that. It's fairly common for people to say "randomly" to mean "uniformly randomly", but it's somewhat sloppy mathematically, and is a terrible thing to do in a probability problem in general. $\endgroup$ – John Hughes Feb 4 at 12:17
  • $\begingroup$ @user376343 I just wanted a hint or clue. Not the whole answer. You would better not taking the matter so serious. It's okay. Your previous answer was enough for me. Because I'm only a 10th grader student. So, I think it is better to avoid the higher knowledge and philosophy for me. $\endgroup$ – Anirban Niloy Feb 4 at 12:31
  • $\begingroup$ @AnirbanNiloy it is not only for you, but for anybody who would later read your question and the answers. I am just cooking something for you :) $\endgroup$ – user376343 Feb 4 at 12:44


On the segment of length $7,$ we are interested of positions of a point $D$ such that the area of the rectangle $DEBF$ is $$\mathcal{A}_{DEBF}>3.$$

The lengths of the sides of $\triangle ABC$ are $$|AB|=7\cdot \sin 15^\circ,\; |BC|=7\cdot \cos 15^\circ.\tag 1$$ Set $\;t=|AD|.$ By similarity of triangles $\triangle ABC \sim \triangle AED$ we get $${|AE|={t\cdot |AB|\over 7}}, \quad |ED|={t\cdot {|BC|}\over 7}.$$ Then $$\mathcal{A}_{DEBF}=|EB|\cdot |ED|=\frac{|AB|\cdot (7-t)}{7}\cdot \frac{|BC|\cdot t}{7}.$$ With the use of $(1)$ we obtain $$\mathcal{A}_{DEBF}=\frac{t(7-t)}{4}$$ which we want larger than $3.$ This asks to solve $$-t^2+7t-12>0,$$ so $t\in(3,4).$ The length of the corresponding segment is $1$ (convenient positions of $D$), the length of $AC$ is $7$ (all possible positions of $D$).

The probability is $1/7.$


2 Let $DC = x , DF = o , DE = a $ and $ DA=y $ .

We have :- $$ o = x \sin 15 \tag{1}$$ $$a= y \ cos 15 \tag{2} $$ Multiplying $(1),(2)$ , we get :- $$ o\cdot a = xy \sin 15 \cos 15 = \frac{xy}{2} sin 30 = \frac{xy}{4} > 3 $$ ( $\because 2 \sin \theta \cos \theta = \sin 2\theta $)

Hence , we need $xy = x(7-x) > 12$ or $x^2-7x+12<0$

As this is of the form $ax^2+bx+c$ , and since $a>0$ , the expression is negative between the roots . Therefore, we must have $x \in (3,4) $

$\therefore $ The probability $\frac{4-3}{7} = \frac{1}{7}$

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    $\begingroup$ @JeanMarie Edited. $\endgroup$ – Sinπ Feb 4 at 12:17
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    $\begingroup$ @JeanMarie There might indeed be some error with my solution ! However , my experiments on GeoGebra reveal that the answer is indeed $\frac{1}{7}$ $\endgroup$ – Sinπ Feb 4 at 12:45
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    $\begingroup$ I am very sorry, you are right. My points are not the good ones. $\endgroup$ – Jean Marie Feb 4 at 12:49

Here is how one can solve the problem with a visualization of the issue, i.e., with a display of the line segment where $D$ must be situated in order that the area condition is fulfilled. Let $(x,y)$ be the coordinates of $D$ with respect to the natural axes of the figure. As the constraint is $xy>3$, the limit is provided by the hyperbola with equation $xy=3$. The coordinates (x_1,y_1) and (x_2,y_2) of intersection points $D_1$ and $D_2$ are thus solutions of the following system :

$$\begin{cases}\frac{x}{s}&+&\frac{y}{s}&=&7\\ &xy&&=&3\end{cases} \ \ \ \text{where} \ s:=\sin(15°) \ \text{and} \ c:=\cos(15°)$$

We have thus transformed our query into a classical problem : we will get a quadratic equation, out of which we will obtain

$$D_1=(\frac{1}{s},\frac{3}{4c}) \ \text{and} \ D_2=(\frac{3}{4s},\frac{1}{c})$$

giving length $D_1D_2=1$. The final answer is $D_1D_2/AC=1/7$ : we find back the same result as @Rahuboy.

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    $\begingroup$ Pardon me. Sorry to say that I'm a student of comparably lower grade and I haven't yet learnt much about hyperbolic function and its property. But thank you for your effort. Hopefully, that suggestion will help me next. $\endgroup$ – Anirban Niloy Feb 4 at 12:03

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