# In a right angled $\triangle ABC$, $DE$ and $DF$ are perpendicular to $AB$ and $BC$ respectively. What is the probability of $DE\cdot DF>3$?

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 $$DE\cdot DF >3?$$

## Attempt:

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.

• 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. – John Hughes Feb 4 at 12:05
• @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. – Anirban Niloy Feb 4 at 12:10
• 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. – John Hughes Feb 4 at 12:17
• @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. – Anirban Niloy Feb 4 at 12:31
• @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 :) – user376343 Feb 4 at 12:44

HINT:

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 $$1/7.$$

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 $$\frac{4-3}{7} = \frac{1}{7}$$

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

• 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. – Anirban Niloy Feb 4 at 12:03