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I am having problems with some problems . I need help. I have tried finding solutions of these, in the forum, still, if there is it, please, share the link.
I am not wishing to repost. So, I will comment with problems, again. I believe, some fellow problem solvers will find these undone, and do them. And, of course, share here with all others including me.



Dhaka 2012 Secondary

$5.$ $AB = 9,$ $AC = 5,$ $BC = 6$ in triangle $ABC$ and the angular bisectors of the three angles are $AD, BF$ and $CE$. Now there is a point in the interior from which the distances of $D, E$ and $F$ are equal. Let this distance be $a$. There’s also another point from which the distances of $A, B$ and $C$ are equal. Let this distance be $b$. Find the area of the rectangle whose two sides are $a$ and $b$.

$6.$Find all solutions for the equation: $(25x^{2}-25)^{2} – (16x^{2}-9)^{2} = (9x^{2}-16)^{2}$



Higher Secondary

$8.$The number $ababab$ has $60$ divisors and the sum of the divisors is $678528$. Find $b/a$.

$10.$Prince charming is outside door $A$ and sleeping beauty is in the grey area. There are $5$ doors and the probabilities of doors $A, B, C, D$ and $E$ being open are $0.8, 0.7, 0.6, 0.5$ and $0.4$. What is the probability of Prince Charming being able to get to sleeping beauty?

6.
$(25x^2-25)^2 - (16x^2-9)^2 = (9x^2-16)^2$
$(25x^2-25+16x^2-9) (25x^2-25-16x^2+9) = (9x^2-16)^2$
$(41x^2-34) (9x^2-16) = (9x^2-16)^2$
$41x^2-34 =9x^2-16$
$32x^2 = 18$
$x^2 = \frac {9}{16}$

$x=+- \frac {3}{4}$

Dhaka 2012 secondary 5
(Calculation brings some results here which I don't want to write so I am only telling how I did it)
First by the well-known theorem that each angle bisector divides its opposite side into segments proportional in length to the adjacent sides,we can get the lengths of $BD,DC,CE,EA,AF,FB$. Now also as we know $AB,BC,CA$ we can determine $\cos A,\cos B,\cos C$. We use it to determine $DE,EF,FD$. Now it is well known that for a triangle with area $\alpha$, circumradius $R$ and side lengths $a,b,c,R=\frac{abc}{4\alpha}$. Using this fact we can determine what we need.

Fahim Shahriar wrote:6.
$(25x^2-25)^2 - (16x^2-9)^2 = (9x^2-16)^2$
$(25x^2-25+16x^2-9) (25x^2-25-16x^2+9) = (9x^2-16)^2$
$(41x^2-34) (9x^2-16) = (9x^2-16)^2$
$41x^2-34 =9x^2-16$
$32x^2 = 18$
$x^2 = \frac {9}{16}$

$x=+- \frac {3}{4}$

You should have considered $9x^{2}-16$ before cancelling from both sides. The given equation has index $4$ so it will have $4$ solutions. If we consider $9x^{2}-16=0$ we have $2$ more solutions and they are $\frac{4}{3}, -\frac{4}{3}$.

For problem 5
I don't know why but I have a strong feeling that what they wanted to define (D,E,F) are touch points of incircle of $\triangle ABC$. Otherwise the calculation will be painful.

sourav das wrote:For problem 5
I don't know why but I have a strong feeling that what they wanted to define (D,E,F) are touch points of incircle of $\triangle ABC$. Otherwise the calculation will be painful.

@Sourav vaia,they told us that $AD,BE,CF$ are angular bisectors,so $D,E,F$ can't be the touch points.
And yeah,assuming them as the intersection points really brings painful calculations.

[quote="SANZEED"]
You should have considered $9x^{2}-16$ before cancelling from both sides. The given equation has index $4$ so it will have $4$ solutions. If we consider $9x^{2}-16=0$ we have $2$ more solutions and they are $\frac{4}{3}, -\frac{4}{3}$. [\quote]
Yes. You're right. That's my most common mistake. Somehow I often do it, even in exams.

$5.$ $AB = 9,$ $AC = 5,$ $BC = 6$ in triangle $ABC$ and the angular bisectors of the three angles are $AD, BF$ and $CE$. Now there is a point in the interior from which the distances of $D, E$ and $F$ are equal. Let this distance be $a$. There’s also another point from which the distances of $A, B$ and $C$ are equal. Let this distance be $b$. Find the area of the rectangle whose two sides are $a$ and $b$.

How can it be "in the interior" while it is an obtuse angled triangle.

Thanks to Sanzeed for reminding me of a theorem named "cosine rule". And Fahim also.

In the probability problem, I have got an assumption. See,if I am correct.

Suppose, "Probability of opened door $A$"= $P_{A}$
then the probability will be
\[P_{A} \cdot \{P_{B}+P_{C}+(P_{D} \cdot P_{E})\}\]

because, all the doors depend on $A$, $E$ depends on $D$
after entering $A$ the prince can get any of the doors $B,C,D$ open. So, they will be added. But, there is another door in $D$. So, it will be multiplied.

Of course, I am not sure

And, please stay tuned. More problems are coming........

SANZEED wrote:

sourav das wrote:For problem 5
I don't know why but I have a strong feeling that what they wanted to define (D,E,F) are touch points of incircle of $\triangle ABC$. Otherwise the calculation will be painful.

@Sourav vaia,they told us that $AD,BE,CF$ are angular bisectors,so $D,E,F$ can't be the touch points.
And yeah,assuming them as the intersection points really brings painful calculations.

I said "wanted to" (may be) . But they didn't. Assuming touch points, it will become a smart divisional point. (Although difficult for those don't know $(ABC)=sr$ )

nafistiham wrote:

$5.$ $AB = 9,$ $AC = 5,$ $BC = 6$ in triangle $ABC$ and the angular bisectors of the three angles are $AD, BF$ and $CE$. Now there is a point in the interior from which the distances of $D, E$ and $F$ are equal. Let this distance be $a$. There’s also another point from which the distances of $A, B$ and $C$ are equal. Let this distance be $b$. Find the area of the rectangle whose two sides are $a$ and $b$.

How can it be "in the interior" while it is an obtuse angled triangle.

Thanks to Sanzeed for reminding me of a theorem named "cosine rule". And Fahim also.

In the probability problem, I have got an assumption. See,if I am correct.

Suppose, "Probability of opened door $A$"= $P_{A}$
then the probability will be
\[$P_{A} \cdot {P_{B}+P_{C}+(P_{D} \cdot P_{E})}$\]

because, all the doors depend on $A$, $E$ depends on $D$
after entering $A$ the prince can get any of the doors $B,C,D$ open. So, they will be added. But, there is another door in $D$. So, it will be multiplied.

Of course, I am not sure

And, please stay tuned. More problems are coming........

If we want to go through door $E$ then $D$ must be opened .
So substitute door $D-E$ with $K$ and the probability would be ($.5 * .4=.2$). So now the total probability would be $.8 \{ \frac{1}{3}(.7+.6+.2) \}$ . (From my point of view as each door has $\frac{1}{3}$ priority)