Problem D and Solution

Many Ways to Get There

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.

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.