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The problem:

In $\triangle PQR$, $ST$ is the perpendicular bisector of $PR$, and $PS$ is the bisector of $\angle QPR$ and the point $S$ lies on $QR$. If $|QS|=9$ and $|SR|=7$ and $|PR|=x/y$, where $x$ and $y$ are co-primes, then what is $x+y$?

I can come up with these steps only:

$\angle STR$ is equal to $90^\circ$, so $\triangle STR$ is a right triangle. The same with $\Delta STP$. In $\triangle PST$ and $\triangle STR$, $ST\cong ST$, $PT\cong TR$, and $\angle PTS= \angle STR$, so that $\triangle PST\cong\triangle STR$. Therefore, $|PS|=|SR|=7$, and $\angle SPT\cong\angle SRT$.

That's the end.

I have no clue how to solve the next steps, or to solve the problem. I don't even know if the above steps lead toward an answer to the problem!

So please help me see how I can solve this step. I am a 8th grader. So please be careful, so that I can understand the steps.

Sorry, I can't add the figure.

Curious learner
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2 Answers2

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$\angle SRP=\angle SPR = \angle SPQ=\theta,$ say. Then adding up the angles in $\triangle PQR$ gives $\angle Q = 180^{\circ}-3\theta,$ and adding up the angles in $\triangle PSQ$ gives $\angle PSQ = 2\theta=\angle QPR.$ That is, $\triangle PQR$ and $\triangle PSQ$ are similar.

Corresponding sides of similar triangles are proportional, so we have $$ \frac{|QP|}{9}=\frac{16}{|QP|}$$ (On the left-hand side we have two sides of $\triangle PQS,$ and on the right-hand side two we have two sides of $\triangle PQR.)$ We conclude that $|QP|=12.$ Now using similarity once again, we have $$\frac{|PR|}{|PQ|}=\frac{7}{9}\implies |PR|=\frac{28}{3}$$

That is, $\boxed{x=28,\space y=3,\space x+y=31}$.

saulspatz
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  • saulspatz, see the edit and help me finding the solution – Curious learner May 11 '18 at 04:29
  • Well if $S$ lies on $QR$, then I've already told you how to do it. – saulspatz May 11 '18 at 04:53
  • I won't understand your answer. I think you done some calculation that is not in grade 8. So, plz help me and edit this as way so I can understand your answer. – Curious learner May 11 '18 at 05:22
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    Do you know that the sum of the angles of a triangle is $180^{\circ}?$ Do you know about similar triangles? Please try to pinpoint what you don't follow in my answer. – saulspatz May 11 '18 at 15:57
  • Yeah, I know these. But I can't catch your calculation. What's mean $\angle Q = 180^{\circ}-3\theta$. that's mean $\angle Q = 180^{\circ}-3\angle SPQ$-what is this---. and also you does some bra bra bra. and in last how can you say that x+y=31, from the similarities. so plz plz edit this Ans with further explanation. – Curious learner May 12 '18 at 01:52
  • saulspatz I tried this your way. But It will be of many values.like:-
    1. $x=77, y=9 \text { so } x+y=86$
    2. $x=70, y=9 \text { so } x+y=79$
    3. many many more***

    Now, there's no need to edit you Ans. But, I wonder that there is no definite value. So which ans. should choose??

     – Curious learner May 12 '18 at 03:28
  • If $S$ lies on $QR$ there is only one solution. I'll try to add some more details to my answer. – saulspatz May 12 '18 at 05:20
  • OK, then add some details – Curious learner May 12 '18 at 11:45
  • Curious learner, Do check your question once. – Love Invariants May 12 '18 at 14:47
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I think this question is contradictory. If you draw a line segment $PR$ and perpendicularly bisect it. Now draw $angle QPR$ with Q as an unfixed point. You will see that the value of $PT,PS,$ $angle QPR$ will remain same but size of $QS$ will differ as $Q$ is an unfixed point along $rayPQ$.
So length of $QS$ is unwanted information.
Hence, it is impossible to find ${x\over y}$.

Love Invariants
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  • I don't quite follow what you say, but I have a sketch that seems to fit the data perfectly. Start with a triangle of sides $12, 16,$ and $28/3.$ Draw the perpendicular bisector of the side of length $28/3.$ If I'm not mistaken, it will divide the side of length $16$ in the ratio $9:7,$ and the point of intersection will lie on the bisector of the opposite angle. – saulspatz May 10 '18 at 17:48
  • saulspatz-Does $S$ lie on $RS$? Otherwise $\angle Q\neq \pi-3\theta$ – Love Invariants May 10 '18 at 18:03
  • I think you mean, does $S$ lie on $QR$? Yes, that is how I interpreted the question, but now I see what you mean; that isn't stated anywhere. It would help to have the figure. – saulspatz May 10 '18 at 18:12