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In triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed. The tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S. Find RS.

 

 Jan 16, 2020
 #1
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-1

RS = 12.

 Jan 17, 2020
 #2
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+2

 

 

THERE WAS AN ERROR SO I DELETED THE PICS.

 Jan 17, 2020
edited by Melody  Jan 17, 2020
edited by Melody  Jan 17, 2020
 #3
avatar+28449 
+2

Deleted as for some reason I used entirely the wrong lengths of triangle legs!

Alan  Jan 17, 2020
edited by Alan  Jan 17, 2020
edited by Alan  Jan 17, 2020
 #5
avatar+28449 
+1

I now think Guest #4 below is correct.  There is a unique answer.  Because that answer is independent of the shape of the triangle it is legitimate to choose an easy one (a right-angled one) on which to do the calculation:

 

RS is indeed 16.

Alan  Jan 17, 2020
 #4
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+1

I think that the answer is unique, and I think that it is 16.

Using Guest's diagram, the triangles QPR and QSP are similar.

The angle between QP and the tangent, call it alpha, is equal to the angle QPS and also the angle QRP.

Similarly the angle between QR and the tangent is equal to the angle QSP and also the angle QPR.

So, QP/QS = QR/QP, 

QS = QP.QP/QR =225/25 = 9, so SR = 16.

 

That fits the approximate values that Melody found by drawing.

 

Alan seems to chosen the particular case where angle PQR is a right angle.

Using the angles I called alpha earlier, from the triangle PQR, tan(alpha) = 15/25 = 3/5.

From the triangle PQS, tan(alpha) = QS/15, so QS = 45/5 = 9, as above.

 Jan 17, 2020
 #6
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Yes you are right guest.

 

I fixed an error in mine and now it always stays at 16.

 

I wss see if I can work out how to give a proper link to what i have created.

 

https://www.geogebra.org/classic/ekztcz7y

 

Melody  Jan 17, 2020
 #7
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Thanks Guest,

 

I am just trying to make sense of your answer.

 

Why is   \(\angle QRP = \alpha\)        ?

Melody  Jan 17, 2020
 #9
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Love the geogebra Melody.

 

In answer to your question, it's a standard angle property of a circle.

Take a diameter from Q across the circle to T on the other side.

Angle QPT will be right angle,since it's based on the diameter, and then angle PTQ will equal the angle between PQ and the tangent, ( subtract from 90 twice). 

Then, all angles based on the same arc, PQ, are equal etc.

Guest Jan 17, 2020
 #11
avatar+107414 
0

Thanks guest for the complement as well as for the explanation.

I am all clear now.

 

Thanks Heureka and Alan as well     laugh laughlaugh

Melody  Jan 18, 2020
 #8
avatar+24031 
+2

In triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed.

The tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S.

Find RS.

\(\text{Let $C$ is the center of the circle } \\ \text{Let $PS=TS=h$ } \\ \text{Let $RS=x$ } \\ \text{Let $QS=25-x$ } \)

 

\(\begin{array}{|lrcll|} \hline 1. & \left(25-x\right)^2+h^2 &=& 15^2 \qquad \text{or}\qquad \mathbf{ h^2 = 15^2-\left(25-x\right)^2} \\\\ 2. & x(25-x) &=& h * h \qquad \text{Intersecting chords theorem} \\ & x(25-x) &=& h^2 \\ & x(25-x) &=& 15^2-\left(25-x\right)^2 \\ & 25x-x^2 &=& 15^2-( 25^2-50x+x^2) \\ & 25x-x^2 &=& 15^2-25^2+50x-x^2 \\ & 25x &=& 15^2-25^2+50x \\ & 25^2-15^2 &=& 25x \\ & 25x &=& 25^2-15^2 \\ & 25x &=& 400 \\ & \mathbf{x} &=& \mathbf{16} \\ \hline \end{array}\)

 

laugh

 Jan 17, 2020
 #10
avatar+24031 
+1

In triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed.
The tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S.
Find RS

 

General solution:

\(\text{Let $Center$ is the center of the circle }\\ \text{Let $P-Center =T-Center=v$ }\\ \text{Let $S-Center =w$ }\\ \text{Let $T-Center =w+k=v\qquad k=v-w$ }\\ \text{Let $ST = k =v-w$ }\\ \text{Let $PS = v+w$ }\\ \text{Let $RS = x$ }\\ \text{Let $QS = 25-x$ }\\ \text{Let $Q-Center = u $ }\)

 

Intersecting chords theorem:

\(\begin{array}{|rcll|} \hline RS*QS &=& ST*PS \\ x*(25-x) &=& (v-w)*(v+w) \\ \mathbf{x*(25-x)} &=& \mathbf{v^2-w^2} & (1) \\ \hline \end{array}\)

 

Pythagoras:

\(\begin{array}{|lrcll|} \hline & v^2+u^2 &=& 15^2 & (2) \\ & w^2+u^2 &=& (25-x)^2 & (3) \\ \hline (2)-(3): & v^2 -w^2 &=& 15^2-(25-x)^2 \quad | \quad \mathbf{x*(25-x)=v^2-w^2} \\ & x*(25-x) &=& 15^2-(25-x)^2 \\ &25x-x^2 &=& 15^2 -(25^2-50x+x^2) \\ &25x-x^2 &=& 15^2 -25^2+50x-x^2 \\ &25x &=& 15^2 -25^2+50x \\ &25^2-15^2 &=& 25x \\ & 25x &=& 25^2-15^2 \\ & 25x &=& 400 \quad |\quad :25 \\ & \mathbf{x} &=& \mathbf{16} \\ \hline \end{array}\)

 

laugh

 Jan 17, 2020

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