Inscribed and Circumscribed Circles, Inscribed Angles: Very Difficult Problems with Solutions

Solution:
The circle through the feet of the altitudes is the nine-point circle; its radius is half the circumradius [tex]R[/tex] of the triangle.
Since [tex]8^2+15^2=17^2[/tex], the triangle formed by the feet is right, so its circumradius is half its hypotenuse: [tex]\frac{17}{2}[/tex].
Hence [tex]R=2 \times \frac{17}{2}=17[/tex].

Solution:
The quadrilateral formed by [tex]C[/tex], [tex]H[/tex] and the feet of the altitudes from [tex]A[/tex] and [tex]B[/tex] has two right angles, so [tex]\angle AHB=180^\circ-\angle ACB=120^\circ[/tex] (vertical angles). If the triangle is obtuse, this angle is [tex]60^\circ[/tex]; the sine is the same in both cases.
By the Law of Sines in [tex]\triangle ABH[/tex]: [tex]R_1=\frac{AB}{2\sin 120^\circ}=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}[/tex].

Solution:
Place the triangle in coordinates: [tex]B(0,0),\ C(14,0),\ A(5,12)[/tex]; then [tex]a=BC=14;\ b=CA=15;\ c=AB=13[/tex].
The incenter is [tex]I=\frac{aA+bB+cC}{a+b+c}=\left(\frac{14 \times 5+13 \times 14}{42},\ \frac{14 \times 12}{42}\right)=(6,4)[/tex].
The circumcenter lies on the perpendicular bisector of [tex]BC[/tex]: [tex]O(7,y)[/tex], and [tex]OB^2=OA^2[/tex] gives [tex]49+y^2=4+(12-y)^2[/tex], so [tex]y=\frac{33}{8}[/tex].
[tex]OI^2=(7-6)^2+\left(\frac{33}{8}-4\right)^2=1+\frac{1}{64}=\frac{65}{64}[/tex], so [tex]OI=\frac{\sqrt{65}}{8}[/tex].

Solution:
The trapezoid is cyclic, so it is isosceles: [tex]AD=BC[/tex]; it is tangential, so [tex]AB+CD=2AD[/tex].
Let [tex]CH=h[/tex] be the altitude from [tex]C[/tex] to [tex]AB[/tex]. Then [tex]AH=\frac{AB+CD}{2}=AD[/tex].
Let [tex]AD=4t[/tex], [tex]AC=5t[/tex]; then [tex]h=\sqrt{AC^2-AH^2}=3t[/tex] and [tex]S=\frac{AB+CD}{2} \times h=4t \times 3t=12t^2=12[/tex], so [tex]t=1[/tex].
Hence [tex]AD=BC=4[/tex], [tex]h=3[/tex], [tex]\frac{AB-CD}{2}=\sqrt{16-9}=\sqrt{7}[/tex], and with [tex]AB+CD=8[/tex]: [tex]AB=4+\sqrt{7};\ CD=4-\sqrt{7}[/tex].

Solution:
Let [tex]CH=h[/tex] be the altitude to the base. The centroid [tex]G[/tex] and the incenter [tex]I[/tex] lie on [tex]CH[/tex], with [tex]GH=\frac{h}{3}[/tex] and [tex]IH=r[/tex].
The incircle meets [tex]CH[/tex] at [tex]H[/tex] and at the point at distance [tex]2r[/tex] from [tex]H[/tex]; since [tex]G \ne H[/tex], we get [tex]GH=2r[/tex], i.e. [tex]h=6r[/tex].
From [tex]S=\frac{ah}{2}=pr[/tex]: [tex]p=\frac{ah}{2r}=3a[/tex], so [tex]P=2p=6a[/tex].

Solution:
Let [tex]H[/tex], [tex]L[/tex], [tex]M[/tex] be their feet on [tex]AB[/tex]. Then [tex]HL=\sqrt{5-1}=2[/tex] and [tex]HM=\sqrt{10-1}=3[/tex].
The bisector from [tex]C[/tex] also bisects the angle between the altitude and the radius to the circumcenter [tex]O[/tex]: [tex]\angle HCL=\angle LCO[/tex].
Use coordinates [tex]H(0,0),\ C(0,1),\ L(2,0),\ M(3,0)[/tex]. Reflecting [tex]\overrightarrow{CH}=(0,-1)[/tex] in the line [tex]CL[/tex] gives the direction [tex](4,3)[/tex] of [tex]\overrightarrow{CO}[/tex], and [tex]O[/tex] lies on the perpendicular bisector of [tex]AB[/tex], the line [tex]x=3[/tex]; so [tex]O\left(3,\frac{13}{4}\right)[/tex] and [tex]R=CO=\frac{15}{4}[/tex].
[tex]\frac{AB}{2}=\sqrt{R^2-\left(\frac{13}{4}\right)^2}=\sqrt{\frac{225-169}{16}}=\frac{\sqrt{14}}{2}[/tex], so [tex]AB=\sqrt{14}[/tex].

Solution:
Each bisector divides [tex]AC[/tex] in the ratio of the adjacent sides, and they meet at the same point, so [tex]\frac{AD}{DC}=\frac{AB}{BC}=\frac{6}{7}[/tex].
By the Law of Cosines [tex]\cos\angle ABC=\frac{66^2+77^2-77^2}{2 \times 66 \times 77}=\frac{3}{7}[/tex]. The quadrilateral is cyclic, so [tex]\cos\angle ADC=-\frac{3}{7}[/tex].
Let [tex]AD=6k;\ DC=7k[/tex]. In [tex]\triangle ACD[/tex]: [tex]77^2=36k^2+49k^2+2 \times 6k \times 7k \times \frac{3}{7}=121k^2[/tex], so [tex]k=7[/tex] and [tex]AD=42;\ CD=49[/tex].

Solution:
Let the side be [tex]a[/tex]. Then [tex]S_{AFB}=\frac{a^2}{10}[/tex], so the distance from [tex]F[/tex] to [tex]AB[/tex] is [tex]h=\frac{a}{5}[/tex].
Put the midpoint of [tex]AB[/tex] at the origin with [tex]AB[/tex] on the x-axis; the center [tex]O[/tex] is at distance [tex]\frac{a}{2}[/tex] on the other side, and the radius squared is [tex]\frac{a^2}{2}[/tex]. If [tex]F[/tex] has abscissa [tex]x[/tex], then [tex]x^2+\left(\frac{a}{2}+\frac{a}{5}\right)^2=\frac{a^2}{2}[/tex], so [tex]x^2=\frac{a^2}{100}[/tex]; since [tex]AF \lt FB[/tex], [tex]x=-\frac{a}{10}[/tex].
[tex]\tan\angle FAB=\frac{h}{x+\frac{a}{2}}=\frac{\frac{a}{5}}{\frac{2a}{5}}=\frac{1}{2}[/tex].

Solution:
Let [tex]r[/tex] be the radius. The height is [tex]2r[/tex], so the leg is [tex]c=\frac{2r}{\sin\alpha}[/tex]; the trapezoid is tangential, so [tex]a+b=2c[/tex] and [tex]S_1=(a+b) \times r=2c \times r=\frac{4r^2}{\sin\alpha}[/tex].
The points of tangency on the bases lie on the axis of symmetry at distance [tex]2r[/tex] from each other; the point of tangency [tex]K[/tex] on a leg is at distance [tex]r\sin\alpha[/tex] from the axis (the radius to [tex]K[/tex] is perpendicular to the leg). So the tangency quadrilateral is a kite with diagonals [tex]2r[/tex] and [tex]2r\sin\alpha[/tex]: [tex]S_2=\frac{1}{2} \times 2r \times 2r\sin\alpha=2r^2\sin\alpha[/tex].
[tex]\frac{S_2}{S_1}=\frac{\sin^2\alpha}{2}=\frac{3}{8}[/tex], so [tex]\sin^2\alpha=\frac{3}{4}[/tex] and [tex]\alpha=60^\circ[/tex].

Solution:
Let [tex]\varphi[/tex] be the inscribed angle subtending each of the three equal chords; then [tex]BC=2R\sin\varphi[/tex], so [tex]\sin\varphi=\frac{1}{2\sqrt{2}}[/tex].
Each of the three equal chords subtends an arc of [tex]2\varphi[/tex], so [tex]\overset{\frown}{AB}=360^\circ-6\varphi[/tex] and [tex]AB=2R\sin(180^\circ-3\varphi)=2R\sin 3\varphi[/tex].
[tex]\sin 3\varphi=3\sin\varphi-4\sin^3\varphi=\frac{3}{2\sqrt{2}}-\frac{1}{4\sqrt{2}}=\frac{5}{4\sqrt{2}}[/tex], so [tex]AB=2 \times 2\sqrt{2} \times \frac{5}{4\sqrt{2}}=5[/tex].

Solution:
Define the point [tex]P[/tex] by [tex]\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}[/tex]. Then [tex]\overrightarrow{AP}=\overrightarrow{OB}+\overrightarrow{OC}=2\overrightarrow{OM}[/tex] and [tex]\overrightarrow{AP}\cdot\overrightarrow{BC}=(\overrightarrow{OB}+\overrightarrow{OC})\cdot(\overrightarrow{OC}-\overrightarrow{OB})=OC^2-OB^2=0[/tex], since [tex]O[/tex] is equally distant from the vertices. So [tex]P[/tex] lies on the altitude from [tex]A[/tex]; in the same way it lies on the altitude from [tex]B[/tex], hence [tex]P=H[/tex].
From [tex]\overrightarrow{AP}=\overrightarrow{OB}+\overrightarrow{OC}=2\overrightarrow{OM}[/tex] we get [tex]AH=2OM[/tex].

Solution:
The tangent segments from the vertex of the right angle to that excircle are equal to the semiperimeter [tex]p[/tex], and together with two radii they form a square, so [tex]r_c=p[/tex]; for the incircle [tex]r=p-c[/tex].
Hence [tex]p=42[/tex] and [tex]c=42-5=37[/tex].
Then [tex]a+b=2p-c=47[/tex] and [tex]ab=2S=2pr=420[/tex], so the legs are the roots of [tex]t^2-47t+420=0[/tex]: [tex]a=35;\ b=12[/tex].

Solution:
By the Law of Sines [tex]a=2R\sin 120^\circ=14[/tex].
The tangent segment from [tex]A[/tex] to the incircle equals [tex]p-a[/tex], so [tex]r=(p-a)\tan\frac{A}{2}[/tex]: [tex]\sqrt{3}=(p-14)\sqrt{3}[/tex], hence [tex]p=15[/tex] and [tex]b+c=16[/tex].
By the Law of Cosines [tex]a^2=b^2+c^2+bc=(b+c)^2-bc[/tex]: [tex]196=256-bc[/tex], so [tex]bc=60[/tex].
Then [tex]b=6;\ c=10[/tex].

Solution:
Place [tex]C(0,0),\ A(b,0),\ B(0,a)[/tex]; then [tex]a^2+b^2=1[/tex]. Let [tex]s=a+b[/tex]; the inradius is [tex]r=\frac{s-1}{2}[/tex] and the incenter is [tex]I(r,r)[/tex]; the centroid is [tex]G\left(\frac{b}{3},\frac{a}{3}\right)[/tex].
The condition [tex]IG=r[/tex] gives [tex]\left(\frac{b}{3}-r\right)^2+\left(\frac{a}{3}-r\right)^2=r^2[/tex], i.e. [tex]\frac{a^2+b^2}{9}-\frac{2rs}{3}+r^2=0[/tex].
Substituting [tex]r=\frac{s-1}{2}[/tex] and multiplying by 36: [tex]4-12s(s-1)+9(s-1)^2=0[/tex], so [tex]3s^2+6s-13=0[/tex] and [tex]s=\frac{-3+4\sqrt{3}}{3}[/tex].
[tex]P=s+1=\frac{4\sqrt{3}}{3}[/tex].

Solution:
[tex]AC[/tex] is a diameter, so [tex]\angle ABC=\angle ADC=90^\circ[/tex]. Let [tex]\angle BAC=x[/tex], [tex]\angle CAD=60^\circ-x[/tex] and [tex]AC=2R[/tex]; then [tex]AB=2R\cos x;\ BC=2R\sin x;\ AD=2R\cos(60^\circ-x);\ CD=2R\sin(60^\circ-x)[/tex].
The quadrilateral is tangential: [tex]AB+CD=BC+AD[/tex], i.e. [tex]\cos x+\sin(60^\circ-x)=\sin x+\cos(60^\circ-x)[/tex], so [tex]\cos x-\sin x=\cos(60^\circ-x)-\sin(60^\circ-x)[/tex]. The function [tex]f(t)=\cos t-\sin t[/tex] is strictly decreasing for these angles, hence [tex]x=60^\circ-x[/tex] and [tex]x=30^\circ[/tex].
So [tex]AB=AD=R\sqrt{3};\ BC=CD=R[/tex], [tex]S=R^2\sqrt{3}[/tex] and [tex]p=R(\sqrt{3}+1)[/tex]. From [tex]S=pr[/tex]: [tex]R^2\sqrt{3}=R(\sqrt{3}+1)[/tex], so [tex]R=\frac{\sqrt{3}+1}{\sqrt{3}}[/tex].
[tex]S=\frac{(\sqrt{3}+1)^2}{\sqrt{3}}=\frac{4+2\sqrt{3}}{\sqrt{3}}=2+\frac{4\sqrt{3}}{3}[/tex].

Solution:
The quadrilateral [tex]AFDC[/tex] is cyclic because [tex]\angle AFC=\angle ADC=90^\circ[/tex]; therefore [tex]\angle BDF=\angle BAC[/tex]. In the same way (with the cyclic quadrilateral [tex]ABDE[/tex]) [tex]\angle CDE=\angle BAC[/tex].
Let [tex]F'[/tex] be the reflection of [tex]F[/tex] in [tex]BC[/tex]. Since [tex]D[/tex] lies on [tex]BC[/tex], [tex]DF'[/tex] is the image of [tex]DF[/tex], and [tex]\angle BDF'=\angle BDF=\angle BAC=\angle CDE[/tex].
So the rays [tex]DF'[/tex] and [tex]DE[/tex] make equal angles with the line [tex]BC[/tex] on opposite sides of [tex]D[/tex], hence [tex]F'[/tex] lies on the extension of [tex]DE[/tex] beyond [tex]D[/tex].
The proof for the reflection in [tex]AC[/tex] is the same.

Solution:
Let [tex]O[/tex] be the center of the circle. For a triangle inscribed in a circle with center [tex]O[/tex], the vector from [tex]O[/tex] to the orthocenter equals the sum of the vectors from [tex]O[/tex] to the vertices (proved in the problem [tex]AH=2OM[/tex] on this page). All four points lie on the same circle, so [tex]\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OD}[/tex] and [tex]\overrightarrow{OQ}=\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OD}[/tex].
Hence [tex]\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=\overrightarrow{OC}-\overrightarrow{OB}=\overrightarrow{BC}[/tex], so [tex]PQ[/tex] is parallel and equal to [tex]BC[/tex].

Solution:
By the Law of Sines [tex]\sin A=\frac{a}{2R}=\frac{8}{17}[/tex], so [tex]\cos A=\pm\frac{15}{17}[/tex]. Also [tex]r=(p-a)\tan\frac{A}{2}[/tex] with [tex]\tan\frac{A}{2}=\frac{\sin A}{1+\cos A}[/tex].
If [tex]\cos A=\frac{15}{17}[/tex]: [tex]\tan\frac{A}{2}=\frac{1}{4}[/tex], so [tex]p-a=24[/tex], [tex]p=40[/tex] and [tex]b+c=64[/tex]; [tex]S=pr=240=\frac{1}{2}bc\sin A[/tex] gives [tex]bc=1020[/tex], and the roots of [tex]t^2-64t+1020=0[/tex] are [tex]b=30;\ c=34[/tex].
If [tex]\cos A=-\frac{15}{17}[/tex]: [tex]\tan\frac{A}{2}=4[/tex], [tex]p=\frac{35}{2}[/tex], [tex]b+c=19[/tex], [tex]bc=\frac{1785}{4}[/tex], and [tex]19^2-4 \times \frac{1785}{4}<0[/tex] — no solution.
The other two sides are [tex]30[/tex] and [tex]34[/tex].

Solution:
[tex]M[/tex] and [tex]N[/tex] lie on the circle with diameter [tex]AB[/tex], so [tex]\angle AMB=\angle ANB=90^\circ[/tex]: they are the feet of the altitudes from [tex]A[/tex] and [tex]B[/tex].
Then [tex]CM=CA\cos C;\ CN=CB\cos C[/tex], so [tex]\triangle CMN \sim \triangle CAB[/tex] and [tex]\frac{MN}{AB}=\frac{CM}{CA}=\cos C[/tex].
[tex]\cos C=\frac{14^2+15^2-13^2}{2 \times 14 \times 15}=\frac{3}{5}[/tex], so [tex]MN=13 \times \frac{3}{5}=\frac{39}{5}[/tex].

Solution:
By the Law of Cosines in [tex]\triangle ABD[/tex] and [tex]\triangle CBD[/tex]: [tex]BD^2=1+36-12\cos A[/tex] and [tex]BD^2=16+9-24\cos C[/tex], where [tex]\cos C=-\cos A[/tex] (the quadrilateral is cyclic).
So [tex]37-12\cos A=25+24\cos A[/tex], i.e. [tex]\cos A=\frac{1}{3}[/tex] and [tex]BD^2=33[/tex].
Then [tex]\sin A=\frac{2\sqrt{2}}{3}[/tex] and, by the Law of Sines in [tex]\triangle ABD[/tex], [tex]R=\frac{BD}{2\sin A}=\frac{3\sqrt{33}}{4\sqrt{2}}=\frac{3\sqrt{66}}{8}[/tex].

Solution:
[tex]\angle A_1JB_1=\angle AJB=180^\circ-\frac{\angle A+\angle B}{2}=90^\circ+\frac{\angle C}{2}[/tex] (vertical angles).
The quadrilateral with vertices [tex]J[/tex], [tex]A_1[/tex], [tex]C[/tex], [tex]B_1[/tex] is cyclic, so its opposite angles are supplementary: [tex]\angle A_1JB_1+\angle C=180^\circ[/tex].
So [tex]90^\circ+\frac{3\angle C}{2}=180^\circ[/tex] and [tex]\angle ACB=60^\circ[/tex].

Solution:
As for any incenter, [tex]\angle AO_1D=90^\circ+\frac{\angle ABD}{2}[/tex] and [tex]\angle AO_2D=90^\circ+\frac{\angle ACD}{2}[/tex]. Since [tex]\angle ABD=\angle ACD[/tex] (inscribed angles on the arc [tex]AD[/tex]), [tex]\angle AO_1D=\angle AO_2D[/tex].
So [tex]O_1[/tex] and [tex]O_2[/tex] see [tex]AD[/tex] at equal angles from the same side, hence [tex]AO_1O_2D[/tex] is cyclic.
[tex]O_1[/tex] lies on the bisector of [tex]\angle BAD[/tex], so [tex]\angle O_1AD=\frac{80^\circ}{2}=40^\circ[/tex], and therefore [tex]\angle O_1O_2D=180^\circ-40^\circ=140^\circ[/tex].

Solution:
[tex]AB=2BC=12[/tex] and [tex]AC=6\sqrt{3}[/tex]. [tex]CL[/tex] bisects the angle at [tex]C[/tex], so [tex]M[/tex] is the midpoint of the arc [tex]\overset{\frown}{AB}[/tex]: [tex]\overset{\frown}{AM}=\overset{\frown}{MB}[/tex]; since [tex]AB[/tex] is a diameter, [tex]AM=BM=\frac{AB}{\sqrt{2}}=6\sqrt{2}[/tex].
By Ptolemy's theorem for the cyclic quadrilateral [tex]ACBM[/tex]: [tex]CM \times AB=AC \times BM+BC \times AM[/tex].
So [tex]CM=\frac{(6\sqrt{3}+6) \times 6\sqrt{2}}{12}=3\sqrt{6}+3\sqrt{2}[/tex].

Solution:
[tex]\angle BAC=\angle CAD=\frac{\alpha}{2}[/tex], and [tex]\angle DCA=\angle BAC=\frac{\alpha}{2}[/tex] (alternate angles), so [tex]AD=DC[/tex].
The trapezoid is cyclic, hence isosceles, and by the Law of Sines [tex]BC=AD=DC=2R\sin\frac{\alpha}{2}[/tex]; also [tex]\angle ACB=180^\circ-\alpha-\frac{\alpha}{2}[/tex], so [tex]AB=2R\sin\angle ACB=2R\sin\frac{3\alpha}{2}[/tex].
The height is [tex]h=BC\sin\alpha[/tex], so [tex]S=\frac{AB+CD}{2} \times h=R\left(\sin\frac{3\alpha}{2}+\sin\frac{\alpha}{2}\right) \times 2R\sin\frac{\alpha}{2}\sin\alpha[/tex].
Since [tex]\sin\frac{3\alpha}{2}+\sin\frac{\alpha}{2}=2\sin\alpha\cos\frac{\alpha}{2}[/tex], [tex]S=4R^2\sin^2\alpha\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}=2R^2\sin^3\alpha[/tex].

Solution:
[tex]M[/tex] and [tex]N[/tex] lie on the circle with diameter [tex]CD[/tex], so [tex]\angle CMD=\angle CND=90^\circ[/tex].
Since [tex]DM \perp BC[/tex] and [tex]CM=MB[/tex], [tex]DM[/tex] is the perpendicular bisector of [tex]BC[/tex], so [tex]DB=DC=L[/tex].
The right triangles [tex]\triangle CMD[/tex] and [tex]\triangle CND[/tex] have the common hypotenuse [tex]CD[/tex] and equal angles at [tex]C[/tex], so [tex]CN=CM[/tex]. Then [tex]AC=\frac{3}{2}CN;\ BC=2CM=2CN[/tex].
By the angle bisector property [tex]\frac{AD}{DB}=\frac{AC}{BC}=\frac{3}{4}[/tex], so [tex]AD=\frac{3L}{4}[/tex] and [tex]AB=AD+DB=\frac{7L}{4}[/tex].

Solution:
Let [tex]O[/tex] be the circumcenter. As proved in the problem [tex]AH=2OM[/tex] on this page, [tex]\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}[/tex].
Then [tex]\overrightarrow{OP}=\frac{\overrightarrow{OA}+\overrightarrow{OB}}{2}[/tex] and [tex]\overrightarrow{OT}=\frac{\overrightarrow{OC}+\overrightarrow{OH}}{2}=\frac{\overrightarrow{OA}+\overrightarrow{OB}}{2}+\overrightarrow{OC}[/tex].
So [tex]\overrightarrow{PT}=\overrightarrow{OT}-\overrightarrow{OP}=\overrightarrow{OC}[/tex], hence [tex]PT=OC=R[/tex].

Solution:
Let [tex]\angle BAD=\alpha[/tex] and [tex]\angle ABC=\beta[/tex], labelled so that [tex]\alpha<\beta[/tex]. The quadrilateral is cyclic, so [tex]\angle BCD=180^\circ-\alpha[/tex] and [tex]\angle ADC=180^\circ-\beta[/tex], and from [tex]\triangle AED[/tex]: [tex]\angle AED=\beta-\alpha[/tex].
Let the bisector of the angle at [tex]E[/tex] meet [tex]BC[/tex] at [tex]K[/tex] and [tex]AD[/tex] at [tex]L[/tex]. From [tex]\triangle EBK[/tex]: [tex]\angle BKE=180^\circ-(180^\circ-\beta)-\frac{\beta-\alpha}{2}=\frac{\alpha+\beta}{2}[/tex]; from [tex]\triangle EAL[/tex]: [tex]\angle ALE=180^\circ-\alpha-\frac{\beta-\alpha}{2}=180^\circ-\frac{\alpha+\beta}{2}[/tex].
Hence [tex]\angle FKL=\angle FLK=\frac{\alpha+\beta}{2}[/tex] (vertical angles at [tex]K[/tex] and supplementary angles at [tex]L[/tex]), so [tex]\triangle FKL[/tex] is isosceles.
In the isosceles [tex]\triangle FKL[/tex] the bisector of the angle at [tex]F[/tex] is perpendicular to the base [tex]KL[/tex], which lies on the bisector of the angle at [tex]E[/tex].

Solution:
As proved in the problem [tex]AH=2OM[/tex] on this page, [tex]\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}[/tex]. Let [tex]\sigma=\overrightarrow{OA}\cdot\overrightarrow{OB}+\overrightarrow{OB}\cdot\overrightarrow{OC}+\overrightarrow{OC}\cdot\overrightarrow{OA}[/tex].
Squaring: [tex]OH^2=3R^2+2\sigma[/tex]. Also [tex]AB^2=|\overrightarrow{OB}-\overrightarrow{OA}|^2=2R^2-2\overrightarrow{OA}\cdot\overrightarrow{OB}[/tex], and similarly for the other sides.
Adding: [tex]AB^2+BC^2+CA^2=6R^2-2\sigma=6R^2-(OH^2-3R^2)=9R^2-OH^2[/tex].

Solution:
Let the legs be [tex]c[/tex]. The trapezoid is tangential, so [tex]a+b=2c[/tex], and the height is [tex]h=2r=c\sin\alpha[/tex].
The projection of a diagonal [tex]d[/tex] onto the longer base is the half-sum of the bases, which equals [tex]c[/tex]: [tex]d^2=h^2+\left(\frac{a+b}{2}\right)^2=h^2+c^2[/tex], so [tex]d=c\sqrt{1+\sin^2\alpha}[/tex].
The circumscribed circle passes through the vertices of [tex]\triangle ABD[/tex], in which the angle [tex]\alpha[/tex] is opposite [tex]d[/tex]: [tex]R=\frac{d}{2\sin\alpha}=\frac{c\sqrt{1+\sin^2\alpha}}{2\sin\alpha}[/tex].
Hence [tex]\frac{r}{R}=\frac{c\sin\alpha}{2} \times \frac{2\sin\alpha}{c\sqrt{1+\sin^2\alpha}}=\frac{\sin^2\alpha}{\sqrt{1+\sin^2\alpha}}[/tex].

Solution:
Take the point [tex]F[/tex] on [tex]DC[/tex] with [tex]DF=DA[/tex] (we assume [tex]F[/tex] lies on [tex]DE[/tex]; the other case is similar). Then [tex]\angle DAF=\angle DFA=\frac{180^\circ-\angle D}{2}=\frac{\angle B}{2}=\angle ABE[/tex].
So [tex]\angle AFE=180^\circ-\frac{\angle B}{2}[/tex] and [tex]\angle AFE+\angle ABE=180^\circ[/tex], hence [tex]ABEF[/tex] is cyclic and [tex]\angle EFB=\angle EAB=\frac{\angle A}{2}[/tex] (inscribed angles on the same arc).
Since [tex]\angle C=180^\circ-\angle A[/tex], in [tex]\triangle FBC[/tex]: [tex]\angle FBC=180^\circ-\angle C-\angle CFB=\frac{\angle A}{2}[/tex], so [tex]CF=CB[/tex].
Therefore [tex]DC=DF+FC=AD+BC[/tex].

Solution:
The trapezoid is isosceles, so both diagonals make the same angle [tex]\theta[/tex] with the longer base; the angle between the diagonals facing a leg is an exterior angle of the triangle formed by them and that base: [tex]2\theta=\varphi[/tex].
So the diagonal is [tex]d=\frac{h}{\sin\theta}=\frac{1}{\sin\frac{\varphi}{2}}[/tex].
The circle is circumscribed about [tex]\triangle ABD[/tex], in which [tex]\alpha[/tex] is opposite [tex]BD[/tex]: [tex]R=\frac{BD}{2\sin\alpha}=\frac{1}{2\sin\alpha\sin\frac{\varphi}{2}}[/tex].

Solution:
The distance from a vertex to the orthocenter is twice the distance from the circumcenter to the opposite side (see the problem [tex]AH=2OM[/tex] on this page), so [tex]AH=2R\cos A[/tex] and [tex]CH=2R\cos C[/tex].
Since [tex]\angle C=180^\circ-2\angle A[/tex], [tex]\cos C=-\cos 2A=1-2\cos^2 A[/tex], and [tex]\cos A=\frac{m}{2R}[/tex].
Then [tex]n=2R\left(1-\frac{2m^2}{4R^2}\right)=2R-\frac{m^2}{R}[/tex], i.e. [tex]2R^2-nR-m^2=0[/tex], and [tex]R=\frac{n+\sqrt{n^2+8m^2}}{4}[/tex].

Solution:
From the altitude from [tex]A[/tex]: [tex]\angle BAH=90^\circ-\angle B[/tex]; from the isosceles [tex]\triangle AOC[/tex]: [tex]\angle OAC=90^\circ-\angle B[/tex]. So [tex]\angle BAH=\angle OAC[/tex], and since [tex]AL[/tex] bisects [tex]\angle BAC[/tex], [tex]\angle LAH=\angle LAO[/tex].
In [tex]\triangle AOH[/tex] the bisector of the angle at [tex]A[/tex] is perpendicular to [tex]OH[/tex], so the triangle is isosceles: [tex]AO=AH[/tex].
Since [tex]AH=2R\cos A[/tex] (see the problem [tex]AH=2OM[/tex] on this page), [tex]R=2R\cos A[/tex], so [tex]\cos A=\frac{1}{2}[/tex] and [tex]\angle BAC=60^\circ[/tex].

Solution:
By Euler's formula the distance between the circumcenter [tex]O[/tex] and the incenter [tex]I[/tex] satisfies [tex]OI^2=R^2-2Rr[/tex].
Since [tex]OI^2 \ge 0[/tex], we get [tex]R^2-2Rr \ge 0[/tex], i.e. [tex]R(R-2r) \ge 0[/tex], and since [tex]R[/tex] is positive, [tex]R \ge 2r[/tex].
Equality [tex]R=2r[/tex] holds exactly when [tex]OI=0[/tex], i.e. when the two centers coincide — for an equilateral triangle.

Solution:
The distance from [tex]O[/tex] to [tex]BC[/tex] is [tex]R\cos A[/tex] and the distance from [tex]I[/tex] to [tex]BC[/tex] is [tex]r[/tex]. Since [tex]OI \parallel BC[/tex], these distances are equal: [tex]R\cos A=r[/tex].
For every triangle [tex]\cos A+\cos B+\cos C=1+\frac{r}{R}[/tex]. Substituting [tex]R\cos A=r[/tex]: [tex]\cos B+\cos C=1[/tex], so [tex]\cos C=1-\frac{\sqrt{2}}{2}[/tex].

Solution:
[tex]\angle B=80^\circ[/tex]. In [tex]\triangle AIC[/tex]: [tex]\angle IAC=30^\circ[/tex] and [tex]\angle AIC=90^\circ+\frac{\angle B}{2}=130^\circ[/tex], so by the Law of Sines [tex]\frac{CI}{\sin 30^\circ}=\frac{AC}{\sin 130^\circ}[/tex], i.e. [tex]CI=\frac{AC}{2\sin 50^\circ}[/tex].
In [tex]\triangle ABC[/tex] by the Law of Sines [tex]\frac{AB}{AC}=\frac{\sin 40^\circ}{\sin 80^\circ}=\frac{\sin 40^\circ}{2\sin 40^\circ\cos 40^\circ}=\frac{1}{2\cos 40^\circ}=\frac{1}{2\sin 50^\circ}[/tex], i.e. [tex]AB=\frac{AC}{2\sin 50^\circ}[/tex].
Hence [tex]CI=AB[/tex].