Some problems of last year divisionals, I need help for
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
- nafistiham
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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?
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?
- Fahim Shahriar
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Re: Some problems of last year divisionals, I need hel for
6.
$(25x^2-25)^2 - (16x^2-9)^2 = (9x^2-16)^2$
$(25x^2-25+16x^2-9) (25x^2-25-16x^2+9) = (9x^2-16)^2$
$(41x^2-34) (9x^2-16) = (9x^2-16)^2$
$41x^2-34 =9x^2-16$
$32x^2 = 18$
$x^2 = \frac {9}{16}$
$x=+- \frac {3}{4}$
$(25x^2-25)^2 - (16x^2-9)^2 = (9x^2-16)^2$
$(25x^2-25+16x^2-9) (25x^2-25-16x^2+9) = (9x^2-16)^2$
$(41x^2-34) (9x^2-16) = (9x^2-16)^2$
$41x^2-34 =9x^2-16$
$32x^2 = 18$
$x^2 = \frac {9}{16}$
$x=+- \frac {3}{4}$
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
Re: Some problems of last year divisionals, I need hel for
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.
(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.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
Re: Some problems of last year divisionals, I need hel for
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 $\frac{4}{3}, -\frac{4}{3}$.Fahim Shahriar wrote:6.
$(25x^2-25)^2 - (16x^2-9)^2 = (9x^2-16)^2$
$(25x^2-25+16x^2-9) (25x^2-25-16x^2+9) = (9x^2-16)^2$
$(41x^2-34) (9x^2-16) = (9x^2-16)^2$
$41x^2-34 =9x^2-16$
$32x^2 = 18$
$x^2 = \frac {9}{16}$
$x=+- \frac {3}{4}$
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
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Re: Some problems of last year divisionals, I need hel for
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.
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.
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
Re: Some problems of last year divisionals, I need hel for
@Sourav vaia,they told us that $AD,BE,CF$ are angular bisectors,so $D,E,F$ can't be the touch points.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.
And yeah,assuming them as the intersection points really brings painful calculations.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Re: Some problems of last year divisionals, I need hel for
[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.
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.
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
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Re: Some problems of last year divisionals, I need hel for
How can it be "in the interior" while it is an obtuse angled triangle.$5.$ $AB = 9,$ $AC = 5,$ $BC = 6$ 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 $b$.
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........
Last edited by nafistiham on Wed Jan 16, 2013 11:00 pm, edited 1 time in total.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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Re: Some problems of last year divisionals, I need hel for
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 $(ABC)=sr$ )SANZEED wrote:@Sourav vaia,they told us that $AD,BE,CF$ are angular bisectors,so $D,E,F$ can't be the touch points.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.
And yeah,assuming them as the intersection points really brings painful calculations.
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
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Re: Some problems of last year divisionals, I need hel for
If we want to go through door $E$ then $D$ must be opened .nafistiham wrote:How can it be "in the interior" while it is an obtuse angled triangle.$5.$ $AB = 9,$ $AC = 5,$ $BC = 6$ 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 $b$.
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........
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)
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )