Trigonometric Geometry

Let \(BD\) be a diameter of the circumcircle. Then \(\angle BCD=90^\circ\), and \(\angle BDC=\angle BAC=A\), since both angles subtend chord \(BC\).

In right triangle \(BCD\), \(BC=BD\sin\angle BDC=2R\sin A\). This proves the claim.

Since triangles \(ABD\) and \(ACD\) have altitudes to the same line \(BC\), \(\frac{S_{ABD}}{S_{ACD}}=\frac{BD}{DC}\).

Using the sine area formula, \(S_{ABD}=\frac12 AB\cdot AD\sin\angle BAD\), \(S_{ACD}=\frac12 AC\cdot AD\sin\angle CAD\). Dividing gives the formula.

Let \(\angle AMB=\theta\). Then \(\angle AMC=180^\circ-\theta\). By the cosine rule, \[ AB^2=AM^2+\frac{a^2}{4}-a\cdot AM\cos\theta, \] \[ AC^2=AM^2+\frac{a^2}{4}+a\cdot AM\cos\theta. \]

Adding gives \(AB^2+AC^2=2AM^2+\frac{a^2}{2}=2AM^2+\frac12 BC^2\).

The area of the whole triangle is \(\frac12 AB\cdot AC\sin A=AB\cdot AC\sin\frac A2\cos\frac A2\).

On the other hand, \(S_{ABC}=S_{ABD}+S_{ACD}=\frac12 AD\cdot AB\sin\frac A2+\frac12 AD\cdot AC\sin\frac A2=\frac12 AD(AB+AC)\sin\frac A2\).

Canceling \(\sin\frac A2\), we get \(AD=\frac{2AB\cdot AC\cos\frac A2}{AB+AC}\).

By trig Ceva, one must check \[ \frac{\sin20^\circ}{\sin40^\circ}\cdot\frac{\sin30^\circ}{\sin50^\circ}\cdot\frac{\sin40^\circ}{\sin30^\circ}=1. \] After canceling \(\sin40^\circ\) and \(\sin30^\circ\), this becomes \(\frac{\sin20^\circ}{\sin50^\circ}\).

Since \(\sin20^\circ\ne\sin50^\circ\), the trig Ceva condition is not satisfied. Therefore concurrence of the cevians does not follow from these data.

By trig Ceva, it is enough to prove \[ \frac{\sin10^\circ}{\sin70^\circ}\cdot \frac{\sin30^\circ}{\sin20^\circ}\cdot \frac{\sin40^\circ}{\sin10^\circ}=1. \] Cancel \(\sin10^\circ\). It remains to prove \(\sin30^\circ\sin40^\circ=\sin20^\circ\sin70^\circ\).

Since \(\sin30^\circ=\frac12\), \(\sin70^\circ=\cos20^\circ\), the right side is \(\sin20^\circ\cos20^\circ=\frac12\sin40^\circ\), equal to the left side.

By Menelaus with directed segments, \[ \frac{BA_1}{CA_1}\cdot\frac{CB_1}{AB_1}\cdot\frac{AC_1}{BC_1}=-1. \]

In each small triangle formed by line \(l\) and two sides of the original triangle, the sine rule expresses a side-segment ratio as a ratio of sines of the angles that \(l\) makes with those sides. Substitution gives the trigonometric condition. Without directed angles, one obtains equality of absolute values to \(1\).

By the cevian formula, \[ \frac{BX}{CX}=\frac{AB\sin\angle BAX}{AC\sin\angle XAC},\quad \frac{BY}{CY}=\frac{AB\sin\angle BAY}{AC\sin\angle YAC}. \] Isogonality gives \(\angle BAX=\angle YAC\), \(\angle XAC=\angle BAY\).

Multiplying cancels the sine factors and leaves \(\frac{AB^2}{AC^2}\).

Let \(D\) be the foot of the altitude from \(A\) to \(BC\). Then \(BD=AB\cos B\) if \(D\) lies on segment \(BC\); in the obtuse case this is interpreted with directed projections.

Similarly, \(DC=AC\cos C\). Adding the projections on line \(BC\), we get \(BC=AB\cos B+AC\cos C\).

Let \(AA_1,BB_1,CC_1\) be the internal angle bisectors. Then \(\angle BAA_1=\angle CAA_1\), so the first trig Ceva factor is \(1\). Similarly, the second and third factors are \(1\).

The product is \(1\), hence by trig Ceva the angle bisectors are concurrent.

For the original cevians, trig Ceva gives \[ X_A X_B X_C=1, \] where \(X_A=\frac{\sin\angle BAA_1}{\sin\angle CAA_1}\), and similarly.

The isogonal cevian at angle \(A\) swaps the two angles with \(AB\) and \(AC\), so the new factor is \(\frac1{X_A}\). Similarly, the others are \(\frac1{X_B}\), \(\frac1{X_C}\). Their product is again \(1\). By trig Ceva, the isogonal cevians are concurrent.

The medians of a triangle are concurrent at the centroid. By the previous problem, the isogonal cevians of a concurrent triple are also concurrent.

The isogonal images of the medians are the symmedians. Therefore the three symmedians are concurrent. Their point is the Lemoine point.

Define rays from \(A,B,C\) so that if they meet at \(T\), the angles around \(T\) become \(120^\circ\). In the standard construction, these are the lines toward the outer vertices of external equilateral triangles.

For these three rays, the angles in trig Ceva are of the form \(B+60^\circ\), \(C+60^\circ\), and cyclically. Substitution gives a telescoping product: \[ \frac{\sin(B+60^\circ)}{\sin(C+60^\circ)} \frac{\sin(C+60^\circ)}{\sin(A+60^\circ)} \frac{\sin(A+60^\circ)}{\sin(B+60^\circ)}=1. \] Hence the rays are concurrent. Their point of concurrence is \(T\), and the construction gives the \(120^\circ\) angles.

Let the outer vertex on side \(BC\) be \(A_1\). From similarity of the constructed isosceles triangles, the direction \(AA_1\) makes angles at \(A\) whose sine ratio is \(\frac{\sin(B+\psi)}{\sin(C+\psi)}\), where \(\psi\) depends only on \(\varphi\).

Cyclically, the three factors are \[ \frac{\sin(B+\psi)}{\sin(C+\psi)} \frac{\sin(C+\psi)}{\sin(A+\psi)} \frac{\sin(A+\psi)}{\sin(B+\psi)}=1. \] By trig Ceva, the lines are concurrent.

Let \(x=\tan\frac A2\), \(y=\tan\frac B2\), \(z=\tan\frac C2\). Since \(\frac A2+\frac B2+\frac C2=90^\circ\), the tangent of the sum is infinite.

Using \(\tan(u+v+w)=\frac{x+y+z-xyz}{1-xy-yz-zx}\), the denominator must be zero. Hence \(1-xy-yz-zx=0\), so \(xy+yz+zx=1\).

For each point on a side, the sine rule in the corresponding small triangle gives equality between the sine ratio and the directed segment ratio on the side.

By assumption, the product of sine ratios is \(-1\), so the product of the corresponding directed segment ratios is also \(-1\). By the converse of Menelaus, \(A_1,B_1,C_1\) are collinear.

Choose a triangle whose sides are three chords of the regular \(18\)-gon so that the given diagonals become cevians. Since the central angle is \(20^\circ\), all angles between chords are half-arc angles, hence multiples of \(10^\circ\).

The angle pairs become \(10^\circ,70^\circ\), \(30^\circ,20^\circ\), \(40^\circ,10^\circ\). By trig Ceva, \[ \frac{\sin10^\circ}{\sin70^\circ}\cdot \frac{\sin30^\circ}{\sin20^\circ}\cdot \frac{\sin40^\circ}{\sin10^\circ}=1. \] This identity follows from \(\sin70^\circ=\cos20^\circ\). Hence the diagonals are concurrent.

Let \(A_h\) be the foot of the altitude from \(A\). If the Euler line is parallel to \(BC\), projecting onto \(AA_h\) and using the centroid ratio on the median gives \(AH:AA_h=2:3\).

We have \(AH=2R\cos A\), and \(AA_h=c\sin B=2R\sin C\sin B\). Hence \[ \frac{\cos A}{\sin B\sin C}=\frac23. \] Substituting \(\cos A=\sin B\sin C-\cos B\cos C\), we obtain \(\sin B\sin C=3\cos B\cos C\), i.e. \(\tan B\tan C=3\). Reversing the steps proves the converse.

Let \(P\) be the first Brocard point, \(\angle ABP=\angle BCP=\angle CAP=\varphi\). Split area \(S\) into \(S_{ABP}+S_{BCP}+S_{CAP}\). Each of these triangles has one angle \(\varphi\), and the neighboring angles are tied to \(A,B,C\).

Using the sine rule in triangles \(ABP\), \(BCP\), \(CAP\), express the ratios \(AP:BP:CP\), then substitute into the area sum. After canceling a common factor, one obtains \[ \operatorname{ctg}\varphi=\operatorname{ctg}A+\operatorname{ctg}B+\operatorname{ctg}C. \] The idea is that the three small areas share \(\sin\varphi\), while the cosine terms collect into the cotangent sum.

In triangle \(AFB\), the points \(X,Y,E\) lie on \(AF,BF,AB\), respectively. Their collinearity is equivalent to Menelaus' condition.

The trigonometric Menelaus form expresses this through the angles made by line \(EXY\) with sides \(AF,BF,AB\). Since \(ABCD\) is cyclic, angles between \(CD\), \(AD\), \(BC\), \(AB\) can be replaced by equal inscribed angles. Thus the entire collinearity check reduces to a product of sines of circle angles.

Use the triangle-angle identity \[ \cos A+\cos B+\cos C=1+4\sin\frac A2\sin\frac B2\sin\frac C2. \] It follows from sum-to-product formulas and \(A+B+C=180^\circ\).

Now apply the standard formula \(\sin\frac A2\sin\frac B2\sin\frac C2=\frac r{4R}\). Hence \(\cos A+\cos B+\cos C=1+\frac rR\).

For the cevian from \(C\) to the vertex of the constructed triangle on \(AB\), the construction angles give an equation of the form \(x\sin(A+\varphi)=y\sin(B+\varphi)\). The other two cevians give cyclic equations.

Solving them gives trilinear coordinates of \(X_\varphi\) reciprocal to \(\sin(A+\varphi),\sin(B+\varphi),\sin(C+\varphi)\), up to a common factor. Isogonal conjugation in trilinears replaces coordinates by reciprocals. Hence the isogonal point has coordinates \((\sin(A+\varphi):\sin(B+\varphi):\sin(C+\varphi))\).

By assumption, the angle order makes trig Ceva reduce to \[ \sin10^\circ\sin20^\circ\sin80^\circ=\sin20^\circ\sin20^\circ\sin30^\circ. \] Cancel \(\sin20^\circ\). We need \(\sin10^\circ\sin80^\circ=\frac12\sin20^\circ\).

Since \(\sin80^\circ=\cos10^\circ\), the left side is \(\sin10^\circ\cos10^\circ=\frac12\sin20^\circ\). Thus trig Ceva holds, and the cevians are concurrent.

Let four lines form four triangles. Choose one triangle and apply trig Menelaus to the remaining line. The product contains sines of angles between the transversal and the three sides of the chosen triangle.

If a neighboring triangle is chosen, one side is replaced by another line of the complete quadrilateral, while the other two remain with opposite orientation. The ratio of the new and old products consists of pairs \(\sin\theta/\sin(180^\circ-\theta)\), each equal to \(1\). Hence the trigonometric Menelaus condition is invariant under the choice of triangle.