difficult geomtrey question?

Question

triangle PQR is similar to triangle PST. Given point P at (0,0), Q at (1,2), R at (2,2), and S at (2,4). Find 2 possible locations at T in exact answers. i know that one point is at (4,4). how do you find the other one in fraction form, without calculator.

I know it is not (2,6)
i think it is a fraction.

it is not (1,6)

i don,t think it is (1/2,7/5) becuase that will have a distance of 2.5 from S

Answer 1

Let's call the two possible values of T: X and Y with X at (4,4). We know that
PS: y = 2x
XY: y = -x/2 + 6. (This follows from point-slope)

Since Y is a reflection of X about line PS, line XY and line PS will intersect at the midpoint of segment XY. We can solve for the coordinates of the midpoint by solving the two eqns above for x and y. Doing so yields (12/5, 24/5) as the midpoint's coordinates. Then to find the coordinates of Y(x,y) we solve the midpoint coordinate eqns:

(1) (x+4)/2 = 12/5
(2) (y+4)/2 = 24/5.

Solving for (x,y) gives (4/5, 28/5)

*Addition: Dr. D's answer check*

I think that your answer check verifies that PRQ is similar to PST but what we want is for PQR to be similar to PST.

The former implies that

Answer 2

You're right that T = (4,4) is a valid point, but you want a different answer.
Consider triangle PST with T presently at (4,4). We want to transform this triangle such that T maps to T' which is identical to S. And S will map to S' which is equivalent to the other value of T you seek.

You'll get (1/2, 7/2)

We need to rotate the original PST about P through an angle arctan(1/3) = arctan(2) - arctan(1) CCW.
Use the matrix:
[3/√10, -1/√10]
[1/√10, 3/√10]

Under this rotation, T goes to (4√10 /5, 8√10 /5)
We need to shrink this to (2,4) to overlap with S.
Multiply it by the factor 5/(2√10). So T' = S.

Now do the same thing to S.
Under the rotation, S goes to (√10 /5, 7√10 /5)
Under the shrinking, it then goes to (1/2, 7/2) or (0.5, 3.5)
So S' = T_alternative = (1/2, 7/2)


*EDIT*
Zenock: please read my answer again. I did not move S. I performed two operations that tranformed S to a new point S', which I'm calling T_alternative. Under these same operations (4,4) is transformed to (2,4) and I'm calling that T' = S. All I did was tranform the triangle, then reset the labeling.

Don't superficially judge an answer. There are ways to check it.

Answer Check:
P = (0,0), R = (2,2), Q = (1,2)
PR = sqrt(8), PQ = sqrt(5), QR = 1

With P = (0,0), S = (2,4), T = (0.5, 3.5)
PS = sqrt(20) = sqrt(8) * sqrt(2.5)
PT = sqrt(12.5) = sqrt(5) * sqrt(2.5)
ST = sqrt(2.5) = 1 * sqrt(2.5)

Clearly these sides are all sqrt(2.5) times larger than the sides of PQR. Thus PST is similar to PQR if T = (0.5, 3.5).

Note: zenock's answer also works, hence there are two solutions. Absird also got the same using a much less cumbersome method than zenock's.

Logic dictates that there should be 3 different solutions since there are 3 ways to orient the triangle while still maintaining similarity. If the question had stated that the two triangles were congruent, then (0.8, 5.6) would have been the only alternative solution.

*EDIT*
zenock wrote: "for PST to be similar to PQR P must correlate to P. S must correlate to Q and T must Correlate to R."

That is wrong. All that needs to happen is for the internal angles to be identical, or for the sides to have the same ratio. It does not matter which point correlates to which.

Absird, there is no difference between triangle PQR and PRQ. They are the same triangle.

Answer 3

(1,6) is not correct. The resulting triangle has lengths PS=2*sqrt(5), ST = sqrt(5), and PT = sqrt(37). This is to be similar to triangle PQR, with lengths PQ = sqrt(5), QR = 1, and PR = 2*sqrt(2). Corresponding sides of similar triangles are in proportion, and there's no way to match up sides of the proposed triangle PST with those of PQR such that the sides are in proportion.

A solution is a triangle that is congruent to the triangle PST that you found with T at (4,4). It's the mirror image of your triangle through PS. A sketch shows that this other location for T has x something less than 1 and y something more than 5.

The exact values are x=0.8 and y=5.6. I got this using some kinda messy algebra. What I did was to define (x,y) to be the coordinates of this other location for T. Then, the square of the distance between P and this point must be equal to the square of the distance between P and your solution (4,4); and the square of the distance between S and this point must equal the square of the distance between S and (4,4). Two equations to be solved simultaneously for the two unknowns x and y.

Is this for an algebra class, or a geometry class? There may be a slicker geometric way to obtain the answer I got.

Answer 4

It's not (1,6)

Distance from (2,4) to (4,4) is 2

Distance from (2,4) to (1,6) is sqroot(5)

Dr. D's answer is incorrect because he moved S. They want the two possible locations of T without moving S

I'll have the correct answer shortly, I'm just trying to find a good way to do it without using a calculator.

Basically you want to find the intersect of the circles...

x^2+y^2=32
and
(x-2)^2+(y-4)^2=4

Graph these circles over your triangles and you will see why.

Basically we the get the equation of the first circle by finding the distance from P to T

and the equation of the second circle by finding the distance from S to T

Because these distances must be the same in the second triangle we know that the intersect of these circles is the two solutions 4,4 being one of them.

Now to solve the equations... here we go, it's a little messy but not too bad...

x^2+y^2=32 (equation 1)
gives us
x^2=32-y^2
and
(x-2)^2+(y-4)^2=4
gives us
x^2-4x+y^2-8y=-16
substitute
32-y^2-4x+y^2-8y=-16
simplify...
4x+8y=48
solve for x
x=12-2y
substitute back into first equation:
(12-2y)^2+y^2=32
simplify:
5y^2-48y+112=0
Use the quadratic equation to solve for y
y=4 and y=5.6

We already have the one at y=4
put 5.6 into first equation and solve for x
x^2+5.6^2=32

x^2=32-31.36=0.64
x=0.8 x=-0.8

Put 5.6 in second equation and solve for x

x^2-4x+5.6^2-8*5.6=-16
simplify
x^2-4x+2.56=0
Solve using the quadratic equation:
x=0.8 and x=2.4

0.8 is the only one that works in both equations so that is the answer.

(0.8,5.6)
Done

Note: Absirds answer is much better. I'm punching myself in the head for not doing it that way, much easier.

I still think Dr. D is misunderstanding the problem. in that for PST to be similar to PQR P must correlate to P. S must correlate to Q and T must Correlate to R. If P and S are fixed and only T can be moved, there is only two possible similar triangles.

Answer 5

I think it is 1,6. Just roll the triangle over and T goes higher.

Answer 6

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Answer 7

its (1,6)