1 Geometry
1.1 Proposals
1.1.1
Proposed by Leon Bankoff. Crux Math. [10: 263].
Show that \[ \frac{\prod\sin A}{\sum\sin A} + \sum\sin^2\frac{A}{2} = 1, \] where the sums and product are cyclic over the angles \(A\), \(B\), \(C\) of a triangle.
1.1.2
Proposed by W. J. Blundon. Amer. Math. Monthly [73: 1122].
If in a triangle \(ABC\) we have \[ \frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C} = \sqrt{3}, \] prove that at least one angle of the triangle is \(60\degree\).
1.1.3
Proposed by Victor Thébault. Amer. Math. Monthly [65: 46].
Let \(O\) be the centre of the circumcircle \((O)\) of a regular convex polygon \(A_1A_2A_3 \dotsm A_{25}A_{26}\) of 26 sides, and let \(O_1\) and \(O_2\) be symmetric to \(O\) with respect to the lines \(A_{25}A_1\) and \(A_2A_6\). Prove that \(O_1O_2\) is equal to a side of an equilateral triangle inscribed in \((O)\).
1.1.4
Proposed by Hippolyte Charles. Crux Math. [5: 199].
Let \(A\), \(B\), \(C\) be the angles of a triangle. Show that \[ \begin{vmatrix} \tan\frac{A}{2} & \cos A & 1 \\ \tan\frac{B}{2} & \cos B & 1 \\ \tan\frac{C}{2} & \cos C & 1 \\ \end{vmatrix} = 0. \]
1.1.5
Proposed by G. C. Giri, Midnapore College. Crux Math. [8: 107].
Prove that, if the incentre \(I\) of a triangle is equidistant from the circumcentre \(O\) and the orthocentre \(H\), then one angle of the triangle is \(60\degree\).
1.1.6
Proposed by W. J. Blundon. Crux Math. [3: 251].
Show how to construct (with a compass and a straight edge) a triangle, given the circumcentre, the incentre and one vertex.
1.1.7
Proposed by Shmuel Avital, I.I.T. Technion. Crux Math. [3: 226].
Find all possible triangles \(ABC\) which have the property that one can draw a line \(AD\), outside the triangular region, on the same side of \(AC\) as \(AB\), which meets \(CB\) (extended) in \(D\) so that triangles \(ABD\) and \(ACD\) will be isosceles.
1.1.8
Proposed by W. J. Blundon. Amer. Math. Monthly [70: 438].
Find necessary and sufficient conditions for a triangle of inradius \(r\), circumradius \(R\), and semiperimeter \(s\) to be isosceles.
1.1.9
Proposed by W. J. Blundon. Crux Math. [3: 42].
Prove that (in the usual notation) the sides of a triangle are in arithmetic progression if and only if \(s^2 = 18Rr - 9r^2\).
Find the corresponding result for geometric progression.
1.1.10
Proposed by W. J. Blundon. Crux Math. [4: 251].
Prove that the line containing the circumcentre and the incentre of a triangle is parallel to a side of the triangle if and only if (in the usual notation) \[ s^2 = \frac{(2R - r)^2 (R + r)}{R - r}. \]
1.1.11
Proposed by Leon Bankoff. Amer. Math. Monthly [61: 476].
Vertices \(A\), \(C\) and \(B\), \(D\) of square \(ABCD\) are joined by quadrants of circles \((B)\) and \((C)\) respectively. A semi-circle \((O_1)\) is described internally on the diameter \(BC\) and a circle \((O_2)\) is drawn tangent to the three arcs. Another circle \((O_3)\) is drawn tangent to circle \((O_2)\) and to arcs \(AC\) and \(BC\), and a right triangle is formed by joining \(O_3\) to \(O_1\) and dropping a perpendicular from \(O_3\) upon \(BC\). Successively tangent circles are drawn in the same manner (with \((O_n)\) tangent to \((O_{n-1})\) and to arcs \(AC\) and \(BC\)) and right triangles are formed (with \(O_1O_n\) for hypotenuse).
Show that the infinitude of triangles so constructed are Pythagorean (integer sides).
Editorial Note. The notation \((X)\) denotes “a circle centered at \(X\)”.
1.1.12
Proposed by W. J. Blundon. Elem. Math. [22: 20].
Let \(I\), \(O\), \(H\) denote respectively the incentre, the circumcentre and the orthocentre of a triangle with sides \(a\), \(b\), \(c\) and the inradius \(r\). Prove that the area \(K\) of the triangle \(IOH\) is given by \[ K = \frac{|(a - b)(b - c)(c - a)|}{8r}. \]
1.1.13
Proposed by George Tsintsifas. Crux Math. [10: 19].
Let \(M\) be any point in the plane of a given triangle \(ABC\). The cevians \(AM\), \(BM\), \(CM\) intersect the lines \(BC\), \(CA\), \(AB\) in \(A^\prime\), \(B^\prime\), \(C^\prime\), respectively. Find the locus of the points \(M\) such that \[ [MCB^\prime] + [MAC^\prime] + [MBA^\prime] = [MC^{\prime}B] + [MA^{\prime}C] + [MB^{\prime}A], \tag{1.1}\] where the square brackets denote the signed area of a triangle.
1.1.14
Proposed W. A. J. Luxemburg, California Institute of Technology. Canad. Math. Bull. [4: 72].
Let \(P_1\), \(P_2\), \(P_3\), \(P_4\) be any four points in the plane, no three collinear. On \(P_iP_{i + 1}\) construct a square with centre \(Q_i\) so that the triangles \(Q_iP_iP_{i + 1}\) all have the same “orientation” (\(i = 1, 2, 3, 4\); \(P_5 = P_1\)). Show that the segments \(Q_1Q_3\) and \(Q_2Q_4\) have the same lengths, and the lines containing them are perpendicular.
1.1.15
Proposed by W. J. Blundon. Crux Math. [7: 179].
Let \(R\), \(r\), \(s\) represent respectively the circumradius, the inradius, and the semiperimeter of a triangle with angles \(\alpha\), \(\beta\), \(\gamma\). It is well-known that \[\begin{align*} \sum \sin \alpha &= \frac{s}{R},\\ \sum \cos \alpha &= \frac{R + r}{R}, \\ \sum \tan \alpha &= \frac{2rs}{s^2 - 4R^2 - 4Rr -r^2}. \end{align*}\]
As for half-angles, it is easy to prove that \(\sum \tan (\alpha / 2) = (4R + r) / s\). Find similar expressions for \(\sum \cos (\alpha / 2)\) and \(\sum \sin (\alpha / 2)\).
1.1.16
Proposed by W. J. Blundon. Crux Math. [3: 155].
Given any triangle (other than equilateral), let \(P\) represent the projection of the incentre \(I\) on the Euler line \(OGNH\) where \(O\), \(G\), \(N\), \(H\) represent respectively the circumcentre, the centroid, the centre of the nine-point circle and the orthocentre of the given triangle. Prove that \(P\) lies between \(G\) and \(H\). In particular, prove that \(P\) coincides with \(N\) if and only if one angle of the given triangle has measure \(60\degree\).
Editorial Comment. See problem 2 above.
1.1.17
Proposed by M. S. Klamkin, University of Alberta. Crux Math. [3: 10].
Let \(P\), \(Q\), \(R\) denote points on the sides \(BC\), \(CA\), and \(AB\), respectively, of a given triangle \(ABC\). Determine all triangles \(ABC\) such that if \[ \frac{BP}{BC} = \frac{CQ}{CA} = \frac{AR}{AB} = k, \quad (\neq 0, 1/2, 1), \] then \(PQR\) (in some order) is similar to \(ABC\).
1.1.18
Proposed by Michael Goldberg. Amer. Math. Monthly [70: 1014].
Find \(r\), the radius of the largest sphere on the surface of which two circles of given radius \(a\), and two circles of given radius \(b\), can be placed so that every point on the surface is within at least one of the regions (less than hemispheres) bounded by the four given circles.
1.2 Solutions
1.2.1
Composite of the solutions of W. J. Blundon, R. H. Eddy, J. Garfunkel, W. Janous, M. S. Klamkin and V. N. Murty. Crux Math. [11: 333].
The proposed equality follows immediately from the following known results, where \(R\), \(r\), \(s\) have their usual meanings: \[ \prod \sin A = \frac{rs}{2R^2}, \quad \sum \sin A = \frac{s}{R}, \quad \sum \sin^2 \frac{A}{2} = 1 - \frac{r}{2R}. \]
1.2.2
Solution by A. Krishnamurthi, Sri Ramakrishna Mission College. Amer. Math. Monthly [75: 405].
We use the following lemma which is easily proved: If \(P + Q + R = 0\), then \[ \sin P + \sin Q + \sin R = -4\sin \frac{P}{2} \sin \frac{Q}{2} \sin \frac{R}{2}. \tag{1.2}\]
Since \(\sqrt{3} = \sin 60\degree / \cos 60\degree\), the given relation is equivalent to \[ \sum (\sin A \cos 60\degree - \cos A \sin 60\degree) = \sum \sin (A - 60\degree) = 0. \]
From (Equation 1.2), \(\prod \sin \frac{1}{2} (A - 60\degree) = 0\), and thus at least one angle must be \(60\degree\).
Editorial Comment. The problem is equivalent to Corollary 1 of Blundon [5]. The corollary states: \[ \text{At least one angle of a triangle is } 60\degree \Longleftrightarrow s = \sqrt{3} (R + r). \]
Since \(\sum \sin A = s/R\) and \(\sum \cos A = (R +r)/R\), the present result follows immediately.
1.2.3
Solution by W. J. Blundon. Amer. Math. Monthly [65: 716].
We may assume the circumcircle has unit radius. Let \(\theta = \pi / 13\), so that \(OO_1 = 2\cos \theta\), \(OO_2 = 2\cos 2\theta\), and angle \(O_1OO_2 = 4\theta\). By the cosine law, \[\begin{align*} O_1O_2^2 &= 4\cos^2 \theta + 4\cos^2 2\theta - 8 \cos \theta \cos 2\theta \cos 4\theta \\ &= 4 + 2\cos 2\theta + 2\cos 4\theta - 8\cos \theta cos 2\theta \cos 4\theta. \end{align*}\]
Upon multiplying by \(\sin \theta\) we have \[\begin{align*} O_1O_2^2 \sin \theta &= 4\sin \theta + (\sin 3\theta - \sin \theta) + (\sin 5\theta - \sin 3\theta) - \sin 8\theta \\ &= 3\sin \theta, \end{align*}\] since \(\sin 8\theta = \sin 5\theta\). It follows that \(O_1O_2 = \sqrt{3}\) which is the length of the side of the inscribed equilateral triangle.
1.2.4
Solution by F. G. B. Maskell, Collège Algonquin. Crux Math. [6: 162-163].
Let \(D\) be the given determinant. We have, all sums being cyclic, \[\begin{align*} D &= \sum \tan \frac{A}{2} (\cos B - \cos C) \\ &= -2\sum \cot \frac{B + C}{2} \sin \frac{B + C}{2} \sin \frac{B - C}{2} \\ &= -2\sum \cos \frac{B + C}{2} \sin \frac{B - C}{2} \\ &= -\sum (\sin B - \sin C) \\ &= 0. \end{align*}\]
Adapted from a comment by W. J. Blundon.
Let \(R\), \(r\) and \(s\) represent respectively the circumradius, inradius, and semiperimeter of a triangle and consider the following two theorems:
Theorem 1. At least one angle of the triangle has measure \(\theta\) if and only if \[ s = 2R\sin \theta + r\cot \frac{\theta}{2}. \tag{1.3}\]
Theorem 2. At least one angle of the triangle has measure \(\theta\) if and only if \[ s\tan \frac{\theta}{2} + 2R\cos \theta = 2R + r. \tag{1.4}\]
It is an easy exercise to show that equations (Equation 1.3) and (Equation 1.4) are equivalent; hence the theorems are also equivalent. The first is a theorem I had proved in 1967 [5].
Now let \(\vec{t}\), \(\vec{c}\), \(\vec{u}\) be the column vectors of the matrix whose determinant is given in the proposal. Since (Equation 1.4) holds when \(\theta\) is replaced by \(A\), \(B\), \(C\) we have \[ s\vec{t} + 2R\vec{c} = (2R + r)\vec{u}. \]
Thus the vectors \(\vec{t}\), \(\vec{c}\), \(\vec{u}\) are linearly dependent and consequently the given determinant vanishes. I had in [5] given a similar corollary to Theorem 1: \[ \begin{vmatrix} 1 & \sin A & \cot \frac{A}{2} \\ 1 & \sin B & \cot \frac{B}{2} \\ 1 & \sin C & \cot \frac{C}{2} \\ \end{vmatrix} = 0. \]
1.2.5
Solution by W. J. Blundon. Crux Math. [9: 153-154].
Let \(\alpha\), \(\beta\), \(\gamma\) be the angles of the triangle. We prove the proposed theorem and its converse: \[ OI = IH \Longleftrightarrow \textrm{at least one of } \alpha, \beta, \gamma\ \textrm{is } 60\degree. \tag{1.5}\]
With the usual meanings for \(R\), \(r\), \(s\), the well-known relations \[ OI^2 = R^2 - 2Rr \quad \textrm{and} \quad IH^2 = 4R^2 + 4Rr + 3r^2 - s^2 \] give \(OI^2 - IH^2 = s^2 - 3(R + r)^2\), and (Equation 1.5) can be replaced by the equivalent \[ s^2 - 3(R + r)^2 = 0 \Longleftrightarrow \text{at least one of } \alpha, \beta, \gamma\ \textrm{is } 60\degree. \tag{1.6}\]
We give two proofs of (Equation 1.6).
Equivalence (Equation 1.6) follows immediately from the known result \[ s^2 - 3(R + r)^2 = R^2 (2\cos \alpha - 1) (2\cos \beta - 1) (2\cos \gamma - 1), \] Crux Math. [4: 59].
Alternatively, equivalence (Equation 1.6) results if we put \(\theta = 60\degree\) in the following known theorem [5]: At least one of the angles of a triangle has measure \(\theta\) if and only if \[ s = 2R\sin \theta + r \cos \frac{\theta}{2}. \]
1.2.6
Solution by Dan Sokolowsky, Antioch College. Crux Math. [4: 136-137].
Let \(O\), \(I\), \(A\) be the three distinct given points which are destined to serve as the circumcentre, incentre, and one vertex of the required triangle (if it exists).
Construction.
Draw circle \((O)OA\) (circle with centre \(O\) and radius \(OA\)). We assume that \(I\) is an interior point of this circle (see Figure Figure 1.1). (Otherwise there is no solution and we are done; for if a solution triangle exists, then \(I\) must be one of its interior points, and hence an interior point of its circumcircle \((O)OA\).) Draw line \(AI\) to meet the circle again in \(T\). Draw the circle \((T)TI\) to meet circle \((O)OA\) in \(B\) and \(C\). Finally, draw \(BC\), \(CA\), \(AB\). Then \(ABC\) is the required triangle.
Proof.
\((\)a\()\) We prove that the triangle \(ABC\) we have constructed is a solution to our problem by showing that \(I\) is its incentre.
Join \(TB\), \(TC\), \(BI\). Since \(TB = TC\), \(AI\) bisects \(\angle A\), and the angles marked \(\alpha\) in the figure are all equal. Let the angles at \(B\) be denoted by \(\beta_1\), \(\beta_2\) as shown; then \[ \alpha + \beta_1 = \angle TIB = \angle TBI = \alpha + \beta_2, \] so \(\beta_1 = \beta_2\), \(BI\) bisects \(\angle B\), and \(I\) is the incentre of triangle \(ABC\).
\((\)b\()\) We prove that our solution is the only one by assuming \(ABC\) to be a solution triangle and showing \(B\) and \(C\) are the intersections of circles \((O)OA\) and \((T)TI\), that is, that \(ABC\) must be the triangle we have in fact constructed. For this, it suffices to show that \(TB = TI = TC\).
We have immediately \(TB = TC\) since \(T\) lies on the bisector of \(\angle A\). Furthermore, we have by assumption \(\beta_1 = \beta_2\), so \[ \angle TIB = \alpha + \beta_1 = \alpha + \beta_2 = \angle TBI; \] hence \(TB = TI\) and the proof is complete.
1.2.7
Solution by W. J. Blundon. Crux Math. [4: 84-85].
Let the angles of \(\triangle ABC\) have measures \(\alpha\), \(\beta\), \(\gamma\). Nine cases are to be considered, according as to which pair of sides are to be equal in \(\triangle ABD\) and which pair in \(\triangle ACD\). Only six of these cases will yield solutions.
Suppose \(ABC\) is a permissible triangle such that \(AB = BD\), and let \(\delta\) be the measure of \(\angle DAB\) (see Figure Figure 1.2). If \(AD = CD\), then \(\gamma < 90\degree\) and \[ \gamma = \alpha + \delta, \quad \beta = 2\delta, \quad \alpha + \beta + \gamma = 180\degree. \]
Eliminating \(\beta\) and \(\delta\) from these equations yields \(\alpha = 3(\gamma - 60\degree)\). Thus \[ \alpha = 3(\gamma - 60\degree), \quad 60\degree < \gamma < 90 \degree \tag{1.7}\] is necessary for the existence of a \(\triangle ABC\) with \(AB = BD\) and \(AD = CD\). Conversely, it is easy to verify that such a triangle can be constructed when (Equation 1.7) holds: hence (Equation 1.7) is necessary and sufficient. A similar analysis would show that for \(AB = BD\) and \(AC = AD\) necessary and sufficient conditions are \[ \alpha = 3(60\degree - \gamma), \quad 0 < \gamma < 60\degree, \] while \(AB = BD\) and \(AC = CD\) is impossible.
Proceeding similarly, we find that a \(\triangle ABC\) exists for which \(AB = AD\) and \(\triangle ACD\) is isoceles if and only if \[ \alpha = \frac{3}{2} (60\degree - \gamma), \quad 0 < \gamma < 60\degree \] or \[ \alpha = 3(60\degree - \gamma), \quad 45\degree < \gamma < 60\degree, \] the lower bound in the second case being determined by the measure of \(\angle DAB\), namely, \(4(\gamma - 45\degree)\); and that one exists for which \(AD = BD\) and \(\triangle ACD\) is isoceles if and only if \[ \alpha = \frac{3}{4} (60\degree - \gamma), \quad 0 < \gamma < 60\degree \] or \[ \alpha = \frac{3}{2} (60\degree - \gamma), \quad 0 < \gamma < 60\degree. \]
Summarizing, we have for \(\triangle ABC\):
no solution for \(\gamma = 60 \degree\) or for \(90\degree \leq \gamma < 180\degree\);
for each \(\gamma\) with \(60\degree < \gamma < 90\degree\), a unique solution \(\alpha = 3(\gamma - 60\degree)\);
for each \(\gamma\) with \(0 < \gamma < 60\degree\), three distinct solutions: \[ \alpha = 3(60\degree - \gamma), \quad \alpha = \frac{3}{2} (60\degree - \gamma), \quad \alpha = \frac{3}{4} (60\degree - \gamma). \]
1.2.8
Solution by D. C. B. Marsh, Colorado School of Mines. Amer. Math. Monthly [71: 213-214].
If we denote the sides of a triangle by \(a\), \(b\), \(c\), we have the easily verified relations \[ a + b + c = 2s, \quad ab + bc + ca = s^2 + 4Rr + r^2, \quad abc = 4Rrs. \]
\(a\), \(b\), \(c\) are the roots of the cubic equation \[ x^3 - 2sx^2 + (s^2 + 4Rr + r^2)x - 4Rrs = 0. \]
the triangle will be isoceles if and only if two roots of this cubic are real and equal; this will be the case if and only if the cubic’s discriminant is zero. The discriminant is \[ 4r^2[4R(R-2r)^3 - (s^2 + r^2 - 10Rr - 2R^2)^2], \] whence a necessary and sufficient condition on \(R\), \(r\), \(s\) that the triangle be isoceles is that \[ 4R(R - 2r)^3 = (s^2 + r^2 -10Rr -2R^2)^2. \]
1.2.9
Solution by W. J. Blundon. Crux Math. [3: 166].
\((\)a\()\) The sides \(a\), \(b\), \(c\) in some order form an arithmetic progression if and only if \[ (2a - b - c) (2b - c - a) (2c - a - b) = 0, \] that is, \((2s - 3a) (2s - 3b) (2s - 3c) = 0\), where \(2s = a + b + c\). This equation reduces to \[ 8s^3 - 12s^2 \sum a + 18s \sum bc - 27abc = 0. \]
By the well-known results \(\sum bc = s^2 + 4Rr + r^2\) and \(abc = 4Rrs\), the equation reduces finally to \[ s^2 = 18Rr - 9 r^2. \]
\((\)b\()\) Here the necessary and sufficient condition is \[ (a^2 - bc) (b^2 - ca) (c^2 - ab) = 0, \] which is equivalent to \((t - a) (t - b) (t - c) = 0\), where \(t^3 = abc\). This equation reduces to \[ t^3 - t^2\sum a + t \sum bc - abc = 0, \] that is \(\sum bc = t \sum a\). Thus \((\sum bc)^3 = abc (\sum a)^3\), whence \[ (s^2 + 4Rr + r^2)^3 = 32Rrs^4. \]
1.2.10
Solution by O. Ramos, Instituto Politécnico José Antonio Echevarría. Crux Math. [5: 201-202].
The line containing the circumcentre and the incentre is parallel to side \(BC\), say, of \(\triangle ABC\) if and only if \(R \cos A = r\). Thus the required condition is equivalent to \[ (R \cos A - r) (R \cos B - r) (R \cos C - r) = 0, \] that is, \[ R^3 \prod \cos A - R^2r \sum \cos B \cos C + Rr^2 \sum \cos A - r^3 = 0. \]
The desired relation now follows upon substituting the well-known relations, Crux Math [4: 59], \[\begin{gather*} \sum \cos A = \frac{R + r}{R}, \\ \sum \cos B \cos C = \frac{s^2 - 4R^2 + r^2}{4R^2}, \\ \prod \cos A = \frac{s^2 - 4R^2 - 4Rr - r^2}{4R^2}. \end{gather*}\]
1.2.11
Solution by W. J. Blundon. Amer. Math. Monthly [62: 734-735].
Let the radius of \((O_n)\) be \(r_n\), let \(P_n\) be the projection of \(O_n\) on \(BC\), and let \(AB = 2\). Then \(O_nB = 2 - r_n\), \(O_1O_n = 1 + r_n\), \(O_1P_n= 1 - 3r_n\), \(O_nP_n = (8r_n - 8r_n^2)^{1/2}\) and \(O_nO_{n + 1} = r_n + r_{n + 1}\).
The relationship \(O_nO_{n + 1}^2 = (O_1P_{n + 1} - O_1P_n)^2 + (O_nP_n - O_{n + 1}P_{n + 1})^2\) gives \[ 9r_{n + 1}^2r_n^2 - 4r_{n + 1}^2r_n - 4r_{n + 1}r_n^2 + 4r_n^2 - 8r_nr_{n + 1} + 4r_{n + 1}^2 = 0. \tag{1.8}\]
Solving this quadratic in \(1/r_{n + 1}\) and selecting the root that gives \(r_{n + 1} < r_n\), we have \[ 1/r_{n + 1} = 1/r_n + 1/2 + (2/r_n - 2)^{1/2}. \tag{1.9}\]
Now from triangle \(O_1O_2B\), \(r_2 = 1/3\) and (Equation 1.9) gives \(r_3 = 2/11\), \(r_4 = 2/18\), \(r_5 = 2/27\), … . The conjecture that \(r_n = 2/(n^2 + 2)\) is easily proved by induction.
Hence the sides of triangle \(O_1O_nP_n\) are found to be \[ \frac{n^2 - 4}{n^2 + 2}, \quad \frac{4n}{n^2 + 2}, \quad \frac{n^2 + 4}{n^2 + 2}. \]
Thus the triangles are all Pythagorean.
It is interesting to note that the only duplicated ratios of sides are \(5:12:13\) corresponding to \(n = 3\) and \(n = 10\), and \(3:4:5\) corresponding to \(n = 4\) and \(n = 6\). Also the centres of the circles lie on the arc of an ellipse with foci \(B\) and \(O_1\).
1.2.12
Solution by O. Reutter. Elem. Math. [23: 16].
Let the vertices of the triangle with respect to circumcentre \(O\) be represented by the vectors \(z_i\). Then, for orthocentre \(H\) and incentre \(I\), the vector representations \(H = \sum z_i\) and \(I = (1/2s) \sum a_iz_i\), are valid, where \(a_i\) is the length of the side opposite the vertex \(z_i\), and \(2s = \sum a_i\). Consequently, triangle \(IOH\) has the area \[ K = \frac{1}{2} \left| \det\left(\sum z_i; \frac{1}{2s} \sum a_iz_i\right) \right| = \frac{1}{4s} \left| \sum (a_{i + 1} - a_i) \det(z_i;z_{i + 1}) \right|. \]
Now, it is well-known that \[ \det (z_i; z_{i + 1}) = \frac{1}{8rs} a_{i + 2}^2 (a_i^2 + a_{i + 1}^2 - a_{i + 2}^2), \] and it follows that \[\begin{align*} K &= \frac{1}{32rs^2} \left| \sum (a_{i + 1} - a_i) a_{i + 2}^2 (a_i^2 + a_{i + 1}^2 - a_{i + 2}^2) \right| \\ &= \frac{1}{32rs^2} \left| \prod (a_{i + 1} - a_i) \cdot (\sum a_i)^2 \right| = \frac{1}{8r} \left| \prod (a_{i + 1} - a_i) \right|, \quad \mathrm{q.e.d.} \\ \end{align*}\]
1.2.13
Solution by W. J. Blundon. Crux Math. [11: 90-91].
In areal coordinates, the coordinates of the vertices of \(ABC\) as triangle of reference, and of an arbitrary point \(M\) in the plane are \[ A(1, 0, 0),\quad B(0, 1, 0),\quad C(0, 0, 1)\quad \mathrm{and}\quad M(x, y, z) \] where \(x + y + z = 1\). Thus the coordinates of \(A^\prime\) are \((0, k, 1 - k)\) for some \(k\), and the collinearity of \(A\), \(M\), \(A^\prime\) gives \[ \begin{vmatrix} 1 & 0 & 0 \\ x & y & z \\ 0 & k & 1 - k \\ \end{vmatrix} = 0 \] whence \[ k = \frac{y}{y + z}. \]
With similar results for \(B^\prime\) and \(C^\prime\) we have \[ A^\prime = \frac{1}{y + z}(0, y, z),\quad B^\prime = \frac{1}{z + x}(x, 0, z),\quad C^\prime = \frac{1}{x + y}(x, y, 0). \]
We now assume without loss of generality that \([ABC] = 1\). Then [6, 228-231] \[\begin{align*} [MCB^\prime] = \frac{1}{z + x} \begin{vmatrix} x & y & z \\ 0 & 0 & 1 \\ x & 0 & z \end{vmatrix} = \frac{xy}{z + x},\\ [MC^{\prime}B] = \frac{1}{x + y} \begin{vmatrix} x & y & z \\ x & y & 0 \\ 0 & 1 & 0 \end{vmatrix} = \frac{zx}{x + y}. \end{align*}\]
With these and similar results, equation (Equation 1.1) in the proposal becomes \[ \frac{xy}{z + x} + \frac{yz}{x + y} + \frac{zx}{y + z} = \frac{zx}{x + y} + \frac{xy}{y + z} + \frac{yz}{z + x}, \] an equation equivalent to \((y - z)(z - x)(x - y)(x + y + z) = 0\), or, since \(x + y + z = 1\), to \((y - z)(z - x)(x - y) = 0\). Thus \(M(x, y, z)\) lies on the locus if and only if at least two of \(x\), \(y\), \(z\) are equal. The required locus is therefore \(l \cap m \cap n\), where \(l\), \(m\), \(n\) are the lines containing the medians through \(A\), \(B\), \(C\), respectively.
1.2.14
Solution by W. J. Blundon. Canad. Math. Bull. [5: 74].
In the complex plane let the number \(2z_k\) be represented by the point \(P_k\). Then with the appropriate “orientation” it is easily verified that the point \(Q_k\) represents the complex number \((1 + i)z_k + (1 - i)z_{k - 1}\). Thus \[ \overline{Q_4Q_2} = (1 + i)(z_2 - z_4) + (1 - i)(z_3 - z_1) = i\overline{Q_3Q_1} \] and the required result follows immediately.
1.2.15
Solution by M. S. Klamkin, University of Alberta. Crux. Math. [8: 188-189].
In an important paper on triangle inequalities [2], Anders Bager tacitly implied (on page 21) that at the present time there are no simple \(R\)-\(r\)-\(s\) representations for the following triangle functions: \[ \sum \sin \frac{\alpha}{2}, \sum \sin \frac{\beta}{2} \sin \frac{\gamma}{2}, \sum \csc \frac{\alpha}{2}, \sum \csc \frac{\beta}{2} \csc \frac{\gamma}{2}; \tag{1.10}\] \[ \sum \cos \frac{\alpha}{2}, \sum \cos \frac{\beta}{2} \cos \frac{\gamma}{2}, \sum \sec \frac{\alpha}{2}, \sum \sec \frac{\beta}{2} \sec \frac{\gamma}{2}. \tag{1.11}\]
However, consideration of \((\sum \sin (\alpha / 2))^2\) shows that, if an \(R\)-\(r\)-\(s\) expression can be found for \(\sum \sin (\alpha / 2)\), then \(R\)-\(r\)-\(s\) expressions can also be found for the remaining functions of (Equation 1.10); and a similar statement can be made for the functions of (Equation 1.11). Now such expressions can be found, but they are not simple, and we conjecture that they cannot be expressed as rational functions of \(R\), \(r\), \(s\). For example, we show how to find an \(R\)-\(r\)-\(s\) expression for \(\sum \cos (\alpha / 2)\); a similar technique will yield an \(R\)-\(r\)-\(s\) expression for \(\sum \sin (\alpha / 2)\). Let \[ t^3 - ut^2 + vt - w = 0 \tag{1.12}\] be the cubic equation whose roots are \(\cos \alpha\), \(\cos \beta\), \(\cos \gamma\). It is known that \[\begin{align*} u &= \sum \cos \alpha = \frac{R + r}{R}, \\ v &= \sum \cos \beta \cos \gamma = \frac{r^2 + s^2 - 4R^2}{4R^2}, \\ w &= \prod \cos \alpha = \frac{s^2 - (2R + r)^2}{4R^2}. \end{align*}\]
With these values of \(u\), \(v\), \(w\), the values of \(\cos \alpha\), \(\cos \beta\), \(\cos \gamma\) can be found from (Equation 1.12), in terms of radicals and substituted in the right member of \[ \sum \cos \frac{\alpha}{2} = \sum \sqrt{(1 + \cos \alpha) / 2}. \]
Editorial Note. W. J. Blundon showed that \(\sum \sin (\alpha / 2)\) is a solution of the quartic equation \[ x^4 = \frac{2R - r}{2R} x^2 + \frac{2r}{R} x + \frac{s^2 - 4R^2 - 4Rr}{4R^2}, \] and that \(\sum \cos (\alpha / 2)\) is a solution of \[ x^4 = \frac{4R + r}{R} x^2 + \frac{2s}{R} x + \frac{s^2}{4R^2}. \]
1.2.16
Solution by W. J. Blundon. Crux Math. [4: 58-59].
To establish the first result, we shall need the following relations, expressed in the usual notation (\(R\), \(r\), \(s\) \(=\) circumradius, inradius, semiperimeter), which are valid for every triangle: \[\begin{align*} GH &= \frac{2}{3} OH, \quad R \geq 2r, \\ IH^2 &= 4R^2 + 4Rr + 3r^2 - s^2, \\ OH^2 &= 9R^2 + 8Rr + 2r^2 - 2s^2, \\ IG^2 &= \frac{1}{9} (s^2 - 16Rr + 5r^2). \end{align*}\]
The first two are well-known, and the remaining ones are easily derivable from known relations. To include isoceles triangles, in which case \(I\) lies on the Euler line, we admit the degenerate triangle \(GIH\).
From the above relations, we have, for nonequilateral triangles, \[ 9(GH^2 - IG^2 - IH^2) = 12r(R - 2r) > 0. \]
This implies that the cosine of angle \(GIH\) is negative, so \(GIH\) is an obtuse (or straight) angle, and \(P\) therefore lies between \(G\) and \(H\).
To prove the second result, we shall use the following additional relations, where \(\alpha\), \(\beta\), \(\gamma\) are the angles of the given triangle: \[\begin{align*} OI^2 &= R^2 - 2Rr, \\ \sum \cos \alpha &= \frac{R + r}{R}, \\ \sum \cos \alpha \cos \beta &= \frac{s^2 - 4R^2 + r^2}{4R^2}, \\ \prod \cos \alpha &= \frac{s^2 - 4R^2 -4Rr - r^2}{4R^2}. \end{align*}\]
The first of these is well-known, and the others are easily derivable from known relations. A straightforward calculation yields \[ (2\cos \alpha - 1) (2\cos \beta - 1) (2\cos \gamma - 1) = \frac{s^2 - 3(R + r)^2}{R^2}, \] which vanishes if and only if \(s = (R + r) \sqrt{3}\). Now, from the well-known relation \(ON = \frac{1}{2} OH\), it follows that \(P\) coincides with \(N\) if and only if \(OI = IH\), that is \[ R^2 - 2Rr = 4R^2 + 4Rr + 3r^2 - s^2, \] which is equivalent to \(s = (R + r) \sqrt{3}\). Hence, \(P\) coincides with \(N\) if and only if one angle of the given triangle has measure \(60 \degree\).
1.2.17
Adapted from solutions submitted independently by Dan Sokolowsky and W. J. Blundon. Crux Math. [3: 160-163].
Consider first the case of scalene triangles and assume without loss of generality that \(a > b > c\). First note that \[ \Delta AQR = \Delta BRP = \Delta CPQ = k(1 - k) \cdot \Delta ABC. \]
Now \(QR\), \(RP\), \(PQ\) are equal in some order to \(\lambda a\), \(\lambda b\), \(\lambda c\) where \[ \lambda ^2 = \frac{\Delta PQR}{\Delta ABC} = 1 - 3k(1 - k) = 1 - 3k + 3k^2 \] and \(k \neq 0, 1/2, 1\).
Apply the cosine law for \(C\) simultaneously in \(\Delta ABC\) and \(\Delta QPC\). Then \[\begin{align*} PQ^2 &= (1 - k^2)^2 a^2 + k^2 b^2 - 2k(1 - k)ab \cdot \frac{a^2 + b^2 - c^2}{2ab} \\ &= (1 - 3k + 2k^2) a^2 + (2k^2 - k)b^2 + (k - k^2)c^2. \end{align*}\]
Let \(PQ = \lambda a\). Then \(PQ^2 - (1 - 3k + 3k^2) a^2 = 0\), that is \[ -k^2 a^2 + (2k^2 - k) b^2 + (k - k^2) c^2 = 0, \] whence \[ k = \frac{b^2 - c^2}{2b^2 - c^2 - a^2}. \]
Similarly, we complete the solution for \(k\) of nine such equations given by one of \(QR\), \(RP\), \(PQ\) \(=\) one of \(\lambda a\), \(\lambda b\), \(\lambda c\) and compile the results, which may be grouped into three cases identified by the subscripts \(i = 1, 2, 3\).
- Case I \[ \frac{P_1Q_1}{b} = \frac{Q_1R_1}{c} = \frac{R_1P_1}{a} \Longleftrightarrow k_1 = \frac{a^2 - b^2}{2a^2 - b^2 - c^2} \quad (0 < k_1 < \frac{1}{2}) \] which has a unique solution for each \(a > b > c\).
- Case II \[ \frac{P_2Q_2}{c} = \frac{Q_2R_2}{a} = \frac{R_2P_2}{b} \Longleftrightarrow k_2 = \frac{c^2 - a^2}{2c^2 - a^2 - b^2} \quad (\frac{1}{2} < k_2 < 1), \] which also has a unique solution for each \(a > b > c\).
- Case III \[ \frac{P_3Q_3}{a} = \frac{Q_3R_3}{b} = \frac{R_3P_3}{c} \Longleftrightarrow k_3 = \frac{b^2 - c^2}{2b^2 - c^2 - a^2}, \] that is, \[ k_3 = \frac{b^2 - c^2}{(b^2 - c^2) - (a^2 - b^2)}. \]
This is impossible if \(0 < k < 1\); but it seems artificial to restrict the points \(P\), \(Q\), \(R\) to lie on the segments \(BC\), \(CA\), \(AB\), and, in any case, the proposal does not require it explicitly. So if we do not impose the restriction \(0 < k < 1\), case III also has a unique solution for each \(a > b > c\).
We have thus shown that for each scalene triangle \(ABC\) there are three distinct triangles \(PQR\) similar to \(ABC\). The similarity in each case is indirect.
The easily proven relation \[ k_{i + 1} = \frac{1 - k_i}{2 - 3k_i}, \quad i = 1,2,3; \quad k_4 = k_1 \] shows how the three solution triangles \(P_iQ_iR_i\) are related to each other.
As an example, Figure 1.4 shows a scalene triangle \(ABC\) and the three triangles \(P_iQ_iR_i\). In that figure \(a = 4\), \(b = 3\), \(c = 2\) and \[\begin{gather*} k_1 = \frac{a^2 - b^2}{2a^2 - b^2 - c^2} = \frac{7}{19}, \quad k_2 = \frac{1 - k_1}{2 - 3k_1} = \frac{12}{17}, \\ k_3 = \frac{1 - k_2}{2 - 3k_2} = - \frac{5}{2}. \end{gather*}\]
Extending the above results to \(a \geq b \geq c\) we have for each equilateral triangle \(a = b = c\) trivially a unique triangle \(PQR\) for every \(k\). Finally, for the sets of isosceles triangles \(a = b > c\) and \(a > b = c\), substitution of \(a = b\) and \(b = c\) in the three cases above gives only the values \(k = 0,1/2,1\), all of which are inadmissable.
Comment by W. J. Blundon. Crux Math. [4: 13].
In a comment on this problem, Crux Math. [3: 197], the editor asked for properties of triangles \(ABC\) with \(a > b > c\) and \(2b^2 = c^2 + a^2\). Because this formula is symmetrical in \(a\) and \(c\), it is sufficient to require \(b\) to be the “middle” side. We will, in a sense, find all triangles such that \[ 2b^2 = c^2 + a^2, \tag{1.13}\] with \(a \geq b \geq c\) or \(a \leq b \leq c\), by assuming \(A\) and \(C\) to be fixed and finding the locus of \(B\). Note that no strictly isosceles (nonequilateral) triangle satisfies (Equation 1.13), for \(b = c\) or \(c = a\) or \(a = b\) all imply \(a = b = c\). Vertices \(A\) and \(C\) being given, let \(O\) be the midpoint of \(AC\) (see Figure 1.5). By a well-known formula for the length of a median, [10] or [11, p. 174], \(ABC\) satisfies (1) if and only if \[ OB = \frac{1}{2} \sqrt{2c^2 + 2a^2 - b^2} = \frac{\sqrt{3}}{2} b. \]
If we draw equitaleral triangle \(ACD\), then \(OD = \frac{\sqrt{3}}{2} b\), and so the locus of \(B\) is the circle with centre \(O\) and radius \(OD\).
1.2.18
Composite of solutions by W. J. Blundon and the proposer. Amer Math. Monthly [71: 1055].
Let the two circles of radius \(a\) intersect at points \(A\) and \(B\). Let the points on the two circles farthest from \(AB\) be \(C\) and \(D\), respectively. Then \(b\) is the circumradius of the triangles \(ACD\) and \(BCD\). Let \(AC = BC = AD = BD = e\). Let \(AB = 2c\), \(CD = 2d\). The situation is, thus, that the four circles intersect in the four vertices of a tetrahedron and, by symmetry, four of its edges are equal. Let \(r\) be the radius of the sphere. Then \[ a = \frac{1}{2} e^2 (e^2 - c^2)^{-1/2}, \quad b = \frac{1}{2} e^2 (e^2 - d^2)^{-1/2}. \]
From this we have \[ 4a^2c^2 = e^2 (4a^2 - e^2), \quad 4b^2d^2 = e^2 (4b^2 - e^2). \]
Let \(h\) be the height of the tetrahedron between the edges \(2c\) and \(2d\). Then, \(h^2 = e^2 - c^2 - d^2\).
If \(f\) is the distance from the centre of the sphere to the midpoint of the edge \(2d\), and \(g\) is the distance to the midpoint of the edge \(2c\), then \(h = f + g\), and \[\begin{align*} r^2 &= f^2 + d^2 = c^2 + g^2 = c^2 + (h - f)^2, \\ f^2 &= c^2 - d^2 + h^2 - 2hf + f^2, \\ f &= (e^2 - 2d^2)/2h, \quad f^2 = (e^2 - 2d^2)^2/4(e^2 - c^2 - d^2), \\ r^2 &= f^2 + d^2 = (e^4 - 4c^2d^2)/4(e^2 - c^2 - d^2) \\ &= e^2(4e^2a^2 + 4e^2b^2 - 12a^2b^2 - e^4)/4(e^2a^2 + e^2b^2 - 4a^2b^2). \end{align*}\]
If the derivative \(dr^2/de = 0\), then \[ (a^2 + b^2)e^6 - [2(a^2 + b^2)^2 + 6a^2b^2]e^4 + 16a^2b^2(a^2 + b^2)e^2 - 24a^4b^4 = 0. \]
This is a cubic in \(e^2\). Solve for \(e^2\) in terms of \(a^2\) and \(b^2\) and substitute in the equation for \(r^2\).
Editorial Note. This is a composite of two independent solutions, one of which presented slightly more motivation for the process and the other slightly more algebraic detail.