# Problem of the Week Problem D and Solution Many Ways to Get There

## Problem

Rectangle $$PQRS$$ has $$QR=4$$ and $$RS=7$$. $$\triangle TRU$$ is inscribed in rectangle $$PQRS$$ with $$T$$ on $$PQ$$ such that $$PT=4$$, and $$U$$ on $$PS$$ such that $$SU=1$$.

Determine the value of $$\angle RUS+\angle PUT$$.

There are many ways to solve this problem. After you have solved it, see if you can solve it a different way.

## Solution

Since $$PQRS$$ is a rectangle, $$PQ=RS$$, so $$TQ=3$$. Similarly $$PS=QR$$, so $$PU=3$$.

We will now present three different solutions. The first uses the Pythagorean Theorem, the second uses congruent triangles, and the third uses basic trigonometry.

Solution 1

Since $$\triangle UPT$$ has a right angle at $$P$$, we can apply the Pythagorean Theorem to find that $$UT^2=PU^2+PT^2 = 3^2 + 4^2 = 25$$. Therefore, $$UT = 5$$, since $$UT >0$$.

Similarly, since $$\triangle TQR$$ has a right angle at $$Q$$, we can apply the Pythagorean Theorem to find that $$TR = 5$$.

Since $$\triangle RSU$$ has a right angle at $$S$$, we can apply the Pythagorean Theorem to find that $$UR^2=RS^2+SU^2= 7^2 + 1^2 = 50$$ and so $$UR = \sqrt{50}$$, since $$UR >0$$.

In $$\triangle TRU$$, notice that $$UT^2+TR^2 = 5^2 + 5^2 = 25 + 25 = 50 = UR^2$$. Therefore, $$\triangle TRU$$ is a right-angled triangle, with $$\angle UTR = 90^{\circ}$$. Also, since $$UT=TR=5$$, $$\triangle TRU$$ is an isosceles right-angled triangle, and so $$\angle TUR = \angle TRU = 45^{\circ}$$.

The angles in a straight line sum to $$180^{\circ}$$, so we have $$\angle RUS + \angle TUR + \angle PUT = 180^{\circ}$$.

Since $$\angle TUR = 45^{\circ}$$, this becomes $$\angle RUS + 45^{\circ} + \angle PUT = 180^{\circ}$$, and so $$\angle RUS + \angle PUT = 180^{\circ} - 45^{\circ} = 135^{\circ}$$. Therefore, $$\angle RUS + \angle PUT = 135^{\circ}$$.

Solution 2

Looking at $$\triangle UPT$$ and $$\triangle TQR$$, we have $$PT = QR=4$$, $$PU=TQ=3$$, and $$\angle UPT = \angle TQR=90^\circ$$. Therefore $$\triangle UPT \cong \triangle TQR$$ by side-angle-side triangle congruency. From the triangle congruency, it follows that $$UT=TR$$, $$\angle QTR = \angle PUT$$, and $$\angle TRQ = \angle PTU$$. Let $$\angle QTR = \angle PUT=x$$ and $$\angle TRQ = \angle PTU=y$$.

Since the angles in a triangle sum to $$180^{\circ}$$, in right-angled $$\triangle UPT$$, $$\angle PUT +\angle PTU=90^\circ$$. That is, $$x+y = 90^{\circ}$$.

Since the angles in a straight line sum to $$180^{\circ}$$, $$\angle PTU + \angle UTR + \angle QTR = 180^{\circ}$$. That is, $$y + \angle UTR + x = 180^{\circ}$$. Substituting $$x+y = 90^{\circ}$$, we obtain $$90^\circ+\angle UTR=180^\circ$$, and $$\angle UTR=90^\circ$$ follows.

Since $$UT=TR$$ and $$\angle UTR=90^\circ$$, $$\triangle TRU$$ is an isosceles right-angled triangle and so $$\angle TUR = \angle TRU = 45^{\circ}$$.

The angles in a straight line sum to $$180^{\circ}$$, so we have $$\angle RUS + \angle TUR + \angle PUT = 180^{\circ}$$.

Since $$\angle TUR = 45^{\circ}$$, this becomes $$\angle RUS + 45^{\circ} + \angle PUT = 180^{\circ}$$, and so $$\angle RUS + \angle PUT = 180^{\circ} - 45^{\circ} = 135^{\circ}$$. Therefore, $$\angle RUS + \angle PUT = 135^{\circ}$$.

Solution 3

Let $$\angle RUS=\alpha$$ and $$\angle PUT=\beta$$.

Using basic trigonometry, from right-angled $$\triangle RSU$$, we have $$\tan \alpha=\frac{7}{1}=7$$, and so $$\alpha=\tan^{-1}(7)$$. Similarly, from right-angled $$\triangle UPT$$, we have $$\tan \beta=\frac{4}{3}$$, and so $$\beta=\tan^{-1}\left(\frac{4}{3}\right)$$.

Then $$\angle RUS + \angle PUT=\alpha +\beta =\tan^{-1}(7)+\tan^{-1}\left(\frac{4}{3}\right)=135^\circ$$.

Therefore, $$\angle RUS + \angle PUT = 135^{\circ}$$.

This third solution is very efficient and concise. However, some of the beauty is lost as a result of this direct approach.