John told his teacher that PQRS is a square, T is any point on SR except S or R, and U is a point on RQ such that angle PTS = angle PTU. Find the measure of angle = TPU.

The measure of angle TPU is 45 degrees.

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Denote the distance from S to T as `x . ` Our goal is to find the angle TPU as a function of `x ` and to prove that it is constant.

Denote also RU as `y .` The picture is attached.

Denote the measure of the angle PTS as...

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Denote the distance from S to T as `x . ` Our goal is to find the angle TPU as a function of `x ` and to prove that it is constant.

Denote also RU as `y .` The picture is attached.

Denote the measure of the angle PTS as `E , ` then the measure of the angle PTU also as `E , ` and the measure of the angle UTR as `pi - 2 E .`

Clearly, `tan E = 1 / x ,` while `tan ( pi - 2 E ) = y / ( 1 - x ) . ` Also,

`tan ( pi - 2 E ) = - tan ( 2 E ) = ( 2 tan E ) / ( tan^2 E - 1 ) = ( 2 / x ) / (1/x^2 - 1) = (2x)/(1-x^2).`

This way, `y = (2x)/(1+x) ` and `1-y=(1-x)/(1+x).`

Now express all sides of the triangle PTU as functions of `x:`

`PT=sqrt(1+x^2), ` `PU = sqrt(1+((1-x)/(1+x))^2), ` `TU = sqrt((1-x)^2+((2x)/(1+x))^2).`

By the Cosine law, the cosine of the angle TPU is `(PT^2+PU^2-TU^2)/(2 PT PU).`

The numerator is `1+2x+((1-x)/(1+x))^2-((2x)/(1+x))^2=`

`= 1 / ( 1 + x )^2 ( ( 1 + 2x ) ( 1 + x )^2 + ( 1 - x )^2-4x^2) = `

`= 1 / ( 1 + x )^2 ( 1+2x+x^2+2x+4x^2+2x^3+1-2x+x^2-4x^2)=` `= 1 / ( 1 + x )^2 (2+2x+2x^2+2x^3)=(2(1+x^2))/(1+x) .`

The denominator is `2 sqrt((1+x^2)(1+((1-x)/(1+x))^2)) =`

`= 2/(1+x) sqrt((1+x^2)(1+x^2+1+x^2))=` ` 2 sqrt(2) (1+x^2) / (1+x) .`

Now we see that the value of `cos (TPU) ` does not depend on `x` and is equal to `1 / sqrt(2), ` so the measure of TPU is always 45 degrees.

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