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

Solution:
The three arcs make up a semicircle: [tex]5x+6x+7x=180^\circ[/tex], so [tex]x=10^\circ[/tex] and [tex]\overset{\frown}{AD}=50^\circ;\ \overset{\frown}{DC}=60^\circ;\ \overset{\frown}{CB}=70^\circ[/tex].
Each angle of [tex]ABCD[/tex] is an inscribed angle, equal to half of the arc it subtends: [tex]\angle A=\frac{60^\circ+70^\circ}{2}=65^\circ[/tex], [tex]\angle B=\frac{50^\circ+60^\circ}{2}=55^\circ[/tex].
[tex]ABCD[/tex] is cyclic, so [tex]\angle C=180^\circ-65^\circ=115^\circ[/tex] and [tex]\angle D=180^\circ-55^\circ=125^\circ[/tex].

Solution:
By the Law of Sines [tex]R=\frac{AB}{2\sin 45^\circ}=\frac{12}{\sqrt{2}}=6\sqrt{2}[/tex].
The arc [tex]AB[/tex] not containing [tex]C[/tex] is twice the inscribed angle: [tex]\overset{\frown}{AB}=2 \times 45^\circ=90^\circ[/tex], so arc [tex]ACB[/tex] measures [tex]360^\circ-90^\circ=270^\circ[/tex].
Its length is [tex]l=\frac{270^\circ}{360^\circ} \times 2\pi \times 6\sqrt{2}=9\sqrt{2}\pi[/tex].

Solution:
Let [tex]AB=18[/tex], [tex]CD=2[/tex]. A trapezoid inscribed in a circle is isosceles: [tex]AD=BC[/tex]. Since a circle can be inscribed, [tex]AD+BC=AB+CD=20[/tex], so [tex]AD=BC=10[/tex].
The projection of a leg onto the longer base is [tex]\frac{18-2}{2}=8[/tex], so the height is [tex]h=\sqrt{10^2-8^2}=6[/tex].
The projection of a diagonal onto the longer base is [tex]\frac{18+2}{2}=10[/tex], so [tex]AC=\sqrt{10^2+6^2}=2\sqrt{34}[/tex] and [tex]AC+BD=4\sqrt{34}[/tex].

Solution:
Let [tex]\angle A=\angle B=30^\circ[/tex], then [tex]\angle C=120^\circ[/tex]. By the Law of Sines [tex]AC=BC=2R\sin 30^\circ=4[/tex] and [tex]AB=2R\sin 120^\circ=4\sqrt{3}[/tex].
[tex]S=\frac{1}{2} \times 4 \times 4 \times \sin 120^\circ=4\sqrt{3}[/tex], [tex]p=\frac{4+4+4\sqrt{3}}{2}=4+2\sqrt{3}[/tex].
[tex]r=\frac{S}{p}=\frac{4\sqrt{3}}{4+2\sqrt{3}}=\frac{2\sqrt{3}}{2+\sqrt{3}}=2\sqrt{3}(2-\sqrt{3})=4\sqrt{3}-6[/tex].

Solution:
Let [tex]CH[/tex] be the altitude to the base, [tex]AH=x[/tex] and [tex]OH=r[/tex] (the incircle touches the base at [tex]H[/tex]). Then [tex]x^2+r^2=45[/tex] and [tex]CH=r+5[/tex].
[tex]\tan\frac{A}{2}=\frac{r}{x}[/tex] and [tex]\tanA=\frac{r+5}{x}[/tex]. Using [tex]\tanA=\frac{2\tan\frac{A}{2}}{1-\tan^2\frac{A}{2}}[/tex] we get [tex](r+5)(x^2-r^2)=2rx^2[/tex].
Substituting [tex]x^2=45-r^2[/tex]: [tex]2r^2+9r-45=0[/tex], so [tex]r=3[/tex] and [tex]x=6[/tex].
Then [tex]CH=8[/tex], [tex]AB=12[/tex] and [tex]AC=BC=\sqrt{6^2+8^2}=10[/tex].

Solution:
The radius to the point of tangency is perpendicular to [tex]BC[/tex], so [tex]OK=r=\sqrt{36-16}=2\sqrt{5}[/tex].
[tex]BO[/tex] and [tex]CO[/tex] bisect the angles of the trapezoid at [tex]B[/tex] and [tex]C[/tex], which add up to [tex]180^\circ[/tex], so [tex]\angle BOC=90^\circ[/tex]. In the right [tex]\triangle BOC[/tex] with altitude [tex]OK[/tex]: [tex]OK^2=BK \times CK[/tex], so [tex]BK=\frac{20}{4}=5[/tex].
By the equal tangent segments [tex]CD=2 \times 4=8[/tex] and [tex]AB=2 \times 5=10[/tex]; the height is [tex]h=2r=4\sqrt{5}[/tex].
[tex]S=\frac{10+8}{2} \times 4\sqrt{5}=36\sqrt{5}[/tex].

Solution:
Let [tex]\triangle ABC[/tex] with [tex]AC=BC[/tex], center [tex]O[/tex] and radius [tex]R[/tex]. The perpendicular from [tex]O[/tex] bisects each chord, so [tex]\frac{AB}{2}=\sqrt{R^2-49}[/tex] and [tex]\frac{BC}{2}=\sqrt{R^2-225}[/tex].
The triangle is acute, so [tex]O[/tex] lies inside it and the altitude to the base is [tex]h=R+7[/tex].
From [tex]BC^2=\left(\frac{AB}{2}\right)^2+h^2[/tex]: [tex]4(R^2-225)=R^2-49+(R+7)^2[/tex], i.e. [tex]R^2-7R-450=0[/tex], so [tex]R=25[/tex].
Then [tex]\frac{AB}{2}=24[/tex], [tex]h=32[/tex] and [tex]S=24 \times 32=768[/tex].

Solution:
By the angle bisector property [tex]AL:LB=AC:BC=36:27=4:3[/tex], so [tex]AL=\frac{4}{7} \times 21=12[/tex].
[tex]AI[/tex] is the bisector of the angle at [tex]A[/tex] in [tex]\triangle ACL[/tex], so [tex]CI:IL=AC:AL=36:12=3:1[/tex].

Solution:
The three triangles are similar: [tex]\triangle ACD \sim \triangle CBD \sim \triangle ABC[/tex], and the radii of their inscribed circles are proportional to their hypotenuses: [tex]\frac{r_1}{AC}=\frac{r_2}{BC}=\frac{r}{AB}[/tex], where [tex]r_1=3[/tex], [tex]r_2=4[/tex].
Squaring and adding, using [tex]AC^2+BC^2=AB^2[/tex]: [tex]r_1^2+r_2^2=r^2[/tex], so [tex]r=\sqrt{9+16}=5[/tex].

Solution:
Let the diagonals meet at [tex]P[/tex], so [tex]\angle APB=90^\circ[/tex], and let [tex]\angle ACB=\varphi[/tex]. Then [tex]\angle AOB=2\varphi[/tex], so the distance from [tex]O[/tex] to [tex]AB[/tex] is [tex]d=R\cos\varphi[/tex].
In the right [tex]\triangle PBC[/tex]: [tex]\angle PBC=90^\circ-\varphi[/tex], and this is the inscribed angle [tex]\angle DBC[/tex] subtending the chord [tex]CD[/tex], so by the Law of Sines [tex]CD=2R\sin(90^\circ-\varphi)=2R\cos\varphi[/tex].
Hence [tex]d=\frac{CD}{2}=5[/tex].

Solution:
[tex]9x=360^\circ[/tex], so [tex]\overset{\frown}{AB}=80^\circ;\ \overset{\frown}{BC}=120^\circ;\ \overset{\frown}{CA}=160^\circ[/tex].
In the quadrilateral formed by the center, two points of tangency and the intersection of their tangents, the angles at the points of tangency are right, so the angle between two tangents equals [tex]180^\circ[/tex] minus the central angle of the arc between the points of tangency.
Hence [tex]\angle M=180^\circ-80^\circ=100^\circ;\ \angle P=180^\circ-120^\circ=60^\circ;\ \angle T=180^\circ-160^\circ=20^\circ[/tex].

Solution:
Let [tex]H[/tex] be the midpoint of the base [tex]AB[/tex]; the incircle touches the base at [tex]H[/tex] and its center lies on the bisector of [tex]\angle BAC[/tex], so [tex]AH=\frac{r}{\tan\frac{A}{2}}=\sqrt{\frac{3}{2}}[/tex], where [tex]\tan\frac{A}{2}=\sqrt{\frac{1-\cos A}{1+\cos A}}=\sqrt{\frac{2}{3}}[/tex].
[tex]\sin A=\sqrt{1-\frac{1}{25}}=\frac{2\sqrt{6}}{5}[/tex], so [tex]\tanA=\frac{\sin A}{\cos A}=2\sqrt{6}[/tex], and the height is [tex]CH=AH \times \tanA=\sqrt{\frac{3}{2}} \times 2\sqrt{6}=6[/tex].
[tex]S=AH \times CH=6\sqrt{\frac{3}{2}}=3\sqrt{6}[/tex].

Solution:
Equal inscribed angles subtend equal arcs: [tex]\overset{\frown}{BC}=\overset{\frown}{CD}[/tex], so [tex]BC=CD[/tex].
The quadrilateral is tangential: [tex]AB+CD=BC+AD[/tex], so [tex]AB=AD[/tex]. Then [tex]\triangle ABC \cong \triangle ADC[/tex] (SSS), so [tex]\angle ABC=\angle ADC[/tex].
The quadrilateral is cyclic: [tex]\angle ABC+\angle ADC=180^\circ[/tex], so [tex]\angle ABC=90^\circ[/tex] and [tex]AC[/tex] is a diameter: [tex]R=\frac{AC}{2}=\frac{a}{2}[/tex].

Solution:
[tex]E[/tex] lies on the circle with diameter [tex]BC[/tex], so [tex]\angle BEC=90^\circ[/tex]. Since [tex]ED \perp BC[/tex], in the right [tex]\triangle BEC[/tex]: [tex]CE^2=CD \times CB[/tex].
From the right [tex]\triangle ADC[/tex]: [tex]CD=AC\cos C[/tex]; since the triangle is isosceles, [tex]\cos C=\frac{AC}{2BC}[/tex].
So [tex]CE^2=AC \times \frac{AC}{2BC} \times BC=\frac{AC^2}{2}=2[/tex], i.e. [tex]CE=\sqrt{2}[/tex].

Solution:
Place the triangle as [tex]A(-\sqrt{3},0),\ B(\sqrt{3},0),\ C(0,1)[/tex]; then [tex]AC=BC=2[/tex].
The orthocenter [tex]H[/tex] lies on the altitude from [tex]C[/tex] (the y-axis): [tex]H(0,y)[/tex], and [tex]\overrightarrow{AH}\cdot\overrightarrow{BC}=0[/tex] gives [tex]\sqrt{3}\cdot(-\sqrt{3})+y\cdot 1=0[/tex], so [tex]y=3[/tex].
The incenter is on the same axis at height [tex]r=\frac{S}{p}=\frac{\sqrt{3}}{2+\sqrt{3}}=2\sqrt{3}-3[/tex]: [tex]I(0,2\sqrt{3}-3)[/tex].
[tex]HI=3-(2\sqrt{3}-3)=6-2\sqrt{3}[/tex].

Solution:
Let [tex]a[/tex], [tex]b[/tex] be the legs and [tex]c[/tex] the hypotenuse. The hypotenuse is a diameter of the circumscribed circle: [tex]c=2R[/tex].
The tangent segments from the vertex of the right angle are equal to [tex]r[/tex], so [tex]a+b=2r+c=2r+2R[/tex].
By the Pythagorean theorem [tex]a^2+b^2=c^2=4R^2[/tex], so [tex]2ab=(a+b)^2-(a^2+b^2)=(2r+2R)^2-4R^2=4r^2+8rR[/tex].
Hence [tex]S=\frac{ab}{2}=r^2+2rR=r(r+2R)[/tex].

Solution:
The quadrilateral [tex]ABCM[/tex] is cyclic, so by Ptolemy's theorem (the product of the diagonals equals the sum of the products of the opposite sides) [tex]MB \times AC=MA \times BC+MC \times AB[/tex].
Since [tex]AB=BC=CA[/tex], dividing by [tex]AC[/tex] gives [tex]MB=MA+MC[/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[/tex] (vertical angles; in an obtuse triangle the angle may also equal [tex]\angle ACB[/tex], and the sine is the same).
Hence [tex]\sin\angle AHB=\sin\angle ACB[/tex], and by the Law of Sines [tex]R_{ABH}=\frac{AB}{2\sin\angle AHB}=\frac{AB}{2\sin\angle ACB}=R[/tex].
In the same way the circumradii of [tex]\triangle BCH[/tex] and [tex]\triangle ACH[/tex] equal [tex]R[/tex].

Solution:
The angle [tex]120^\circ[/tex] is the vertex angle. Let [tex]AC=BC=b[/tex]; then [tex]AB=b\sqrt{3}[/tex].
[tex]S=\frac{1}{2}b^2\sin 120^\circ=\frac{\sqrt{3}}{4}b^2[/tex], [tex]p=\frac{(2+\sqrt{3})b}{2}[/tex], so [tex]r=\frac{S}{p}=\frac{\sqrt{3}\,b}{2(2+\sqrt{3})}=1[/tex], i.e. [tex]b=\frac{2(2+\sqrt{3})}{\sqrt{3}}[/tex].
[tex]R=\frac{AB}{2\sin 120^\circ}=\frac{b\sqrt{3}}{\sqrt{3}}=b[/tex], so [tex]R=\frac{4+2\sqrt{3}}{\sqrt{3}}=2+\frac{4\sqrt{3}}{3}[/tex].

Solution:
By the Law of Sines [tex]\sin\angle ACB=\frac{AB}{2R}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}[/tex]; the triangle is acute, so [tex]\angle ACB=60^\circ[/tex].
Since [tex]\overset{\frown}{AD}=\overset{\frown}{DB}[/tex], the inscribed angles on these arcs are equal, so [tex]CD[/tex] bisects the angle at [tex]C[/tex]: [tex]\angle ACD=\frac{\angle ACB}{2}=30^\circ[/tex].

Solution:
Let [tex]AB=6[/tex], [tex]AC=BC=9[/tex], and let the incircle touch [tex]AC[/tex] and [tex]BC[/tex] at [tex]K[/tex] and [tex]L[/tex]. With [tex]p=\frac{6+9+9}{2}=12[/tex], the tangent segments from [tex]C[/tex] are [tex]CK=CL=p-AB=6[/tex].
By symmetry [tex]KL \parallel AB[/tex], so [tex]\frac{KL}{AB}=\frac{CK}{CA}[/tex], i.e. [tex]KL=6 \times \frac{6}{9}=4[/tex].

Solution:
The height equals the diameter: [tex]h=2r=6[/tex]. The legs are [tex]c_1=\frac{6}{\sin 60^\circ}=4\sqrt{3}[/tex] and [tex]c_2=\frac{6}{\sin 45^\circ}=6\sqrt{2}[/tex].
The trapezoid is tangential, so [tex]a+b=c_1+c_2[/tex] and [tex]P=2(c_1+c_2)=8\sqrt{3}+12\sqrt{2}[/tex].

Solution:
Equal chords subtend equal arcs: [tex]\overset{\frown}{BC}=\overset{\frown}{CD}[/tex], so [tex]\angle CAD=\angle CDB[/tex] (inscribed angles on these arcs).
The triangles also share [tex]\angle C[/tex], so [tex]\triangle CDE \sim \triangle CAD[/tex] and [tex]\frac{CD}{CA}=\frac{CE}{CD}[/tex].
Hence [tex]CA=\frac{CD^2}{CE}=\frac{36}{3}=12[/tex] and [tex]P=12+6+7=25[/tex].

Solution:
In the rhombus [tex]PA=PB=OA=R[/tex], so [tex]\triangle AOP[/tex] is equilateral and [tex]\angle AOP=60^\circ[/tex].
The diagonal of a rhombus bisects its angle, so [tex]\angle AOB=120^\circ[/tex]; this central angle subtends the arc containing [tex]P[/tex], and [tex]C[/tex] lies on the other arc.
Hence [tex]\angle ACB=\frac{120^\circ}{2}=60^\circ[/tex].

Solution:
The trapezoid is tangential, so [tex]a+b=2c[/tex]; hence the leg equals the midsegment: [tex]m=\frac{a+b}{2}=c=80[/tex].
[tex]h=c\sin 30^\circ=40[/tex], and the height is the diameter: [tex]r=\frac{h}{2}=20[/tex].

Solution:
By the Law of Cosines [tex]BC^2=48+36-2 \times 4\sqrt{3} \times 6 \times \frac{\sqrt{3}}{2}=12[/tex], so [tex]BC=2\sqrt{3}[/tex].
[tex]AC^2+BC^2=48=AB^2[/tex], so [tex]\angle ACB=90^\circ[/tex] and [tex]\angle ABC=60^\circ[/tex].
The trapezoid is cyclic, hence isosceles: [tex]AD=BC[/tex], and [tex]CD=AB-2BC\cos 60^\circ=4\sqrt{3}-2\sqrt{3}=2\sqrt{3}[/tex].

Solution:
A diameter perpendicular to a chord bisects the arcs cut off by the chord, so [tex]\overset{\frown}{CB}=\overset{\frown}{BD}[/tex].
The inscribed angles [tex]\angle CPB[/tex] and [tex]\angle BPD[/tex] subtend these equal arcs, so [tex]\angle CPB=\angle BPD[/tex], i.e. [tex]PB[/tex] is the bisector of [tex]\angle CPD[/tex].

Solution:
[tex]AB[/tex] is a diameter, so [tex]\angle ACB=90^\circ[/tex] and [tex]AC=\sqrt{36-6}=\sqrt{30}[/tex].
The altitude from [tex]C[/tex] is [tex]h=\frac{AC \times BC}{AB}=\frac{\sqrt{180}}{6}=\sqrt{5}[/tex], and its foot [tex]H[/tex] gives [tex]BH=\frac{BC^2}{AB}=1[/tex].
The trapezoid is isosceles, so [tex]CD=AB-2BH=4[/tex] and [tex]S=\frac{6+4}{2} \times \sqrt{5}=5\sqrt{5}[/tex].

Solution:
[tex]p=\frac{9}{2}[/tex], so the tangent segments from [tex]A[/tex] are [tex]AM=AN=p-BC=\frac{5}{2}[/tex].
By the Law of Cosines [tex]\cos A=\frac{16+9-4}{2 \times 4 \times 3}=\frac{7}{8}[/tex], so [tex]\sin A=\frac{\sqrt{15}}{8}[/tex].
[tex]S=\frac{1}{2} \times \frac{5}{2} \times \frac{5}{2} \times \frac{\sqrt{15}}{8}=\frac{25\sqrt{15}}{64}[/tex].

Solution:
[tex]OB[/tex] and [tex]OC[/tex] bisect the exterior angles at [tex]B[/tex] and [tex]C[/tex], so [tex]\angle OBC=\frac{180^\circ-\angle B}{2}[/tex] and [tex]\angle OCB=\frac{180^\circ-\angle C}{2}[/tex].
Hence [tex]\angle BOC=180^\circ-\angle OBC-\angle OCB=\frac{\angle B+\angle C}{2}=90^\circ-\frac{\angle A}{2}[/tex].
So [tex]\frac{\angle A}{2}=50^\circ[/tex] and [tex]\angle BAC=100^\circ[/tex].

Solution:
The leg perpendicular to the bases is the height: [tex]AD=h=2r=2[/tex].
[tex]BC=\frac{h}{\sin 30^\circ}=4[/tex].
The trapezoid is tangential, so [tex]AB+CD=AD+BC=6[/tex] and [tex]P=2 \times 6=12[/tex].

Solution:
[tex]c=\sqrt{15^2+20^2}=25[/tex] and [tex]r=\frac{15+20-25}{2}=5[/tex].
Place [tex]C(0,0),\ A(20,0),\ B(0,15)[/tex]; then [tex]I(5,5)[/tex]. The hypotenuse lies on [tex]3x+4y=60[/tex], and the altitude [tex]CH[/tex] is perpendicular to it through the origin: [tex]4x-3y=0[/tex].
[tex]d=\frac{|4 \times 5-3 \times 5|}{5}=1[/tex].

Solution:
[tex]c=2R[/tex] and [tex]r=\frac{a+b-c}{2}[/tex]; with [tex]a=c\sin A;\ b=c\cos A[/tex] we get [tex]\frac{R}{r}=\frac{c}{a+b-c}=\frac{1}{\sin A+\cos A-1}[/tex].
Hence [tex]\sin A+\cos A=1+\frac{1}{\sqrt{3}+1}=\frac{1+\sqrt{3}}{2}[/tex], and since [tex]\sin A+\cos A=\sqrt{2}\sin(A+45^\circ)[/tex], [tex]\sin(A+45^\circ)=\frac{1+\sqrt{3}}{2\sqrt{2}}=\sin 75^\circ[/tex].
So [tex]A=30^\circ[/tex] or [tex]A=60^\circ[/tex]: the acute angles are [tex]30^\circ[/tex] and [tex]60^\circ[/tex].

Solution:
Let [tex]AC=BC[/tex], [tex]CH=30[/tex], with [tex]H[/tex] on the base. The centroid [tex]G[/tex] and the circumcenter [tex]O[/tex] lie on [tex]CH[/tex], and [tex]GH=10[/tex].
Since [tex]OG=5[/tex], either [tex]OH=15[/tex] or [tex]OH=5[/tex], and [tex]R=CO=30-OH[/tex].
If [tex]OH=15[/tex], then [tex]R=15[/tex] and [tex]\frac{AB}{2}=\sqrt{R^2-OH^2}=0[/tex] — impossible. So [tex]OH=5[/tex], [tex]R=25[/tex] and [tex]\frac{AB}{2}=\sqrt{625-25}=10\sqrt{6}[/tex].

Solution:
Both circles touch the diagonal [tex]BD[/tex]; since they touch each other, they touch [tex]BD[/tex] at the same point [tex]K[/tex].
The tangent segment from [tex]B[/tex] gives [tex]BK=\frac{AB+BD-AD}{2}[/tex] in the first triangle and [tex]BK=\frac{BC+BD-CD}{2}[/tex] in the second.
Hence [tex]AB-AD=BC-CD[/tex], i.e. [tex]AB+CD=AD+BC=20[/tex], so [tex]P=40[/tex].

Solution:
The trapezoid is tangential: [tex]AB+CD=AD+BC=18[/tex]. Let [tex]AD=h[/tex]; then [tex]BC=18-h[/tex].
The projection of [tex]BC[/tex] onto the longer base is [tex]12-6=6[/tex], so [tex]BC^2=h^2+6^2[/tex], i.e. [tex](18-h)^2=h^2+36[/tex], hence [tex]h=8[/tex] and [tex]BC=10[/tex].
The obtuse and the acute angle at the leg [tex]BC[/tex] are supplementary, so [tex]\sin\angle BCD=\sin\angle ABC=\frac{h}{BC}=\frac{4}{5}[/tex].

Solution:
From the right [tex]\triangle ACH[/tex]: [tex]\sin A=\frac{CH}{AC}=\frac{5}{6}[/tex].
By the Law of Sines [tex]2R=\frac{BC}{\sin A}=\frac{15 \times 6}{5}=18[/tex]. (In general [tex]2R=\frac{AC \times BC}{CH}[/tex].)

Solution:
[tex]h=2r=4[/tex]; the projection of a leg onto the longer base is [tex]\frac{a-b}{2}=\frac{h}{\tan\alpha}=1[/tex], so the leg is [tex]c=\sqrt{1+16}=\sqrt{17}[/tex].
The trapezoid is tangential, so [tex]a+b=2c=2\sqrt{17}[/tex]. Hence [tex]a=\sqrt{17}+1;\ b=\sqrt{17}-1[/tex].
[tex]S=\frac{a+b}{2} \times h=4\sqrt{17}[/tex].

Solution:
The segment [tex]CD[/tex] is seen from [tex]A[/tex] and [tex]B[/tex] at equal angles ([tex]\angle DAC=\angle DBC=50^\circ[/tex]) from the same side, so [tex]ABCD[/tex] is cyclic.
Inscribed angles on the same arc are equal: [tex]\angle ACB=\angle ADB=45^\circ[/tex].
Then [tex]\angle ACD=100^\circ-45^\circ=55^\circ[/tex] and [tex]\angle ABD=\angle ACD=55^\circ[/tex].

Solution:
The trapezoid is isosceles; the projection of a leg onto the longer base is [tex]\frac{8-2}{2}=3[/tex], so [tex]h=\sqrt{16-9}=\sqrt{7}[/tex].
The projection of a diagonal onto the longer base is the half-sum of the bases, so [tex]BD=\sqrt{h^2+\left(\frac{8+2}{2}\right)^2}=\sqrt{32}=4\sqrt{2}[/tex], and [tex]\sin A=\frac{h}{AD}=\frac{\sqrt{7}}{4}[/tex].
The circle is circumscribed about [tex]\triangle ABD[/tex], so by the Law of Sines [tex]R=\frac{BD}{2\sin A}=\frac{4\sqrt{2}}{\frac{\sqrt{7}}{2}}=\frac{8\sqrt{14}}{7}[/tex].

Solution:
Let [tex]x[/tex] be the tangent segment from the third vertex. The sides are [tex]6+8=14[/tex], [tex]6+x[/tex] and [tex]8+x[/tex], and [tex]p=14+x[/tex].
[tex]S=pr=4(14+x)[/tex], and by Heron's formula, where [tex]p-a;\ p-b;\ p-c[/tex] are the tangent segments, [tex]S=\sqrt{p \times x \times 6 \times 8}[/tex].
So [tex]16(14+x)^2=48x(14+x)[/tex], i.e. [tex]16(14+x)=48x[/tex], hence [tex]x=7[/tex] and the other sides are [tex]13;\ 15[/tex].

Solution:
Let the legs be [tex]c[/tex]. The trapezoid is tangential: [tex]a+b=2c[/tex]; the projections of the legs give [tex]a-b=2c\cos\alpha[/tex].
Adding: [tex]c=\frac{a}{1+\cos\alpha}[/tex], so [tex]b=a-2c\cos\alpha=\frac{a(1-\cos\alpha)}{1+\cos\alpha}=a\tan^2\frac{\alpha}{2}[/tex].
The height equals the diameter, so [tex]r=\frac{h}{2}=\frac{c\sin\alpha}{2}=\frac{a\sin\alpha}{2(1+\cos\alpha)}=\frac{a}{2}\tan\frac{\alpha}{2}[/tex], where [tex]\tan\frac{\alpha}{2}=\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}[/tex].

Solution:
From [tex]OA \times OC=OB \times OD[/tex]: [tex]\frac{OA}{OD}=\frac{OB}{OC}[/tex]. Also [tex]\angle AOB=\angle DOC[/tex] (vertical angles), so [tex]\triangle AOB \sim \triangle DOC[/tex] (SAS similarity).
Hence [tex]\angle OAB=\angle ODC[/tex], i.e. [tex]\angle CAB=\angle CDB[/tex].
So the segment [tex]BC[/tex] is seen from [tex]A[/tex] and [tex]D[/tex] at equal angles from the same side, and [tex]ABCD[/tex] is cyclic.

Solution:
The incenter lies on the angle bisectors of [tex]ABCD[/tex]; since [tex]O[/tex] lies on [tex]AC[/tex], the diagonal [tex]AC[/tex] bisects the angles at [tex]A[/tex] and [tex]C[/tex]: [tex]\angle BAC=\angle DAC[/tex] and [tex]\angle BCA=\angle DCA[/tex].
Then [tex]\triangle ABC \cong \triangle ADC[/tex] (common side [tex]AC[/tex]), so [tex]AB=AD;\ CB=CD[/tex].
In the same way [tex]BD[/tex] bisects the angles at [tex]B[/tex] and [tex]D[/tex], so [tex]BA=BC[/tex].
Hence [tex]AB=BC=CD=DA[/tex], and [tex]ABCD[/tex] is a rhombus.

Solution:
With area [tex]S[/tex] and semiperimeter [tex]p[/tex]: [tex]r=\frac{S}{p}[/tex] and [tex]r_a=\frac{S}{p-a};\ r_b=\frac{S}{p-b};\ r_c=\frac{S}{p-c}[/tex].
Then [tex]\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}=\frac{(p-a)+(p-b)+(p-c)}{S}=\frac{3p-2p}{S}=\frac{p}{S}=\frac{1}{r}[/tex].

Solution:
In any right triangle [tex]r=\frac{a+b-c}{2}[/tex], where [tex]a[/tex] and [tex]b[/tex] are the legs and [tex]c[/tex] is the hypotenuse. All three triangles are right, so [tex]r_1=\frac{AC+BC-AB}{2}[/tex], [tex]r_2=\frac{AP+CP-AC}{2}[/tex] and [tex]r_3=\frac{BP+CP-BC}{2}[/tex].
Adding and using [tex]AP+BP=AB[/tex]: [tex]r_1+r_2+r_3=\frac{AP+BP+2CP-AB}{2}=CP[/tex].

Solution:
Let [tex]a=BC[/tex], [tex]b=CA[/tex], and [tex]p=\frac{a+b+c}{2}[/tex]. The tangent segments give [tex]AM=p-a[/tex] for the incircle and [tex]AN=p-b[/tex] for the excircle.
Both points lie on [tex]AB[/tex], so [tex]MN=|AM-AN|=|b-a|[/tex].

Solution:
The trapezoid is cyclic, hence isosceles: [tex]AC=BD[/tex] and [tex]\angle ABC=\alpha[/tex]. Let the diagonals meet at [tex]P[/tex]; [tex]\triangle ABP[/tex] is isosceles and right, so [tex]\angle PAB=\angle PBA=45^\circ[/tex].
In [tex]\triangle ABC[/tex]: [tex]\angle ACB=180^\circ-45^\circ-\alpha=135^\circ-\alpha[/tex].
By the Law of Sines [tex]AB=2R\sin(135^\circ-\alpha)=2R\sin(45^\circ+\alpha)[/tex].

Solution:
Place [tex]A(-b,0),\ B(b,0),\ C(0,h)[/tex]. Then [tex]H\left(0,\frac{b^2}{h}\right)[/tex], and the incenter lies on [tex]CM[/tex] at height [tex]r=\frac{bh}{b+\sqrt{b^2+h^2}}[/tex].
The incircle meets the axis [tex]CM[/tex] at the foot of the altitude and at height [tex]2r[/tex], so [tex]\frac{b^2}{h}=2r[/tex], i.e. [tex]b\sqrt{b^2+h^2}=2h^2-b^2[/tex].
Squaring: [tex]b^2(b^2+h^2)=(2h^2-b^2)^2[/tex], hence [tex]5b^2=4h^2[/tex] and [tex]h=\frac{\sqrt{5}}{2}b[/tex].
Then [tex]HM=\frac{b^2}{h}=\frac{2b}{\sqrt{5}};\ CH=h-HM=\frac{b}{2\sqrt{5}}[/tex], so [tex]CH:HM=1:4[/tex].

Solution:
The trapezoid is isosceles; let [tex]AB=a[/tex], [tex]CD=b[/tex]. The incircle touches the bases at their midpoints, so the tangent segments give [tex]AK=\frac{a}{2};\ DK=\frac{b}{2}[/tex], and [tex]AD=\frac{a+b}{2}[/tex]. By symmetry [tex]KL \parallel AB[/tex], and [tex]K[/tex] divides [tex]AD[/tex] in the ratio [tex]a:b[/tex], so it lies at the fraction [tex]\frac{a}{a+b}[/tex] of the height [tex]h[/tex] measured from [tex]AB[/tex].
Since [tex]\triangle AOB[/tex] and [tex]\triangle COD[/tex] are similar, [tex]\frac{AO}{OC}=\frac{AB}{CD}=\frac{a}{b}[/tex]; hence [tex]O[/tex] lies on the axis of symmetry at the same fraction [tex]\frac{a}{a+b}[/tex] of the height measured from [tex]AB[/tex].
So [tex]O[/tex] lies on the line [tex]KL[/tex].

Solution:
Let the bases be [tex]a[/tex] and [tex]b[/tex]. The trapezoid is tangential, so the leg equals the midsegment: [tex]c=m=5[/tex], and [tex]a+b=2m=10[/tex].
Each part has half the height of the trapezoid: [tex]S_1=\frac{b+5}{2} \times \frac{h}{2};\ S_2=\frac{a+5}{2} \times \frac{h}{2}[/tex], so [tex](b+5):(a+5)=7:13[/tex]. Since [tex](b+5)+(a+5)=20[/tex], we get [tex]b=2[/tex] and [tex]a=8[/tex].
The projection of a leg is [tex]\frac{a-b}{2}=3[/tex], so [tex]h=\sqrt{25-9}=4[/tex] and [tex]r=2[/tex].

Solution:
The incircle touches the bases at their midpoints, and the tangent segments from a vertex are equal: [tex]\frac{CD}{2}=CM=3[/tex] and [tex]\frac{AB}{2}=MB=6[/tex], so [tex]CD=6[/tex], [tex]AB=12[/tex] and [tex]BC=9[/tex].
The projection of the leg onto the longer base is [tex]\frac{12-6}{2}=3[/tex], so [tex]h=\sqrt{9^2-3^2}=6\sqrt{2}[/tex].
The projection of the diagonal is the half-sum of the bases, so [tex]AC=\sqrt{h^2+\left(\frac{AB+CD}{2}\right)^2}=\sqrt{72+81}=3\sqrt{17}[/tex].

Solution:
By the power of the point [tex]C[/tex] with respect to the circle (the lines [tex]CA[/tex] and [tex]CB[/tex] meet it at [tex]M[/tex], [tex]A[/tex] and at [tex]N[/tex], [tex]B[/tex]): [tex]CM \times CA=CN \times CB[/tex], i.e. [tex]\frac{CA^2}{2}=\frac{CB^2}{2}[/tex], so [tex]CA=CB[/tex]; since [tex]AB=2[/tex], [tex]CA=CB=\sqrt{2}[/tex].
Place [tex]C(0,0),\ A(\sqrt{2},0),\ B(0,\sqrt{2})[/tex]. The center [tex]Q[/tex] lies on the perpendicular bisectors of [tex]AM[/tex] and [tex]BN[/tex]: [tex]Q\left(\frac{3\sqrt{2}}{4},\frac{3\sqrt{2}}{4}\right)[/tex].
[tex]R^2=QA^2=\left(\frac{\sqrt{2}}{4}\right)^2+\left(\frac{3\sqrt{2}}{4}\right)^2=\frac{5}{4}[/tex], so [tex]R=\frac{\sqrt{5}}{2}[/tex].

Solution:
From [tex]S=\frac{a h_a}{2}[/tex] and the similar formulas: [tex]a=\frac{2S}{h_a};\ b=\frac{2S}{h_b};\ c=\frac{2S}{h_c}[/tex].
Then [tex]S=pr=\frac{a+b+c}{2}r=S\left(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right)r[/tex], so [tex]\frac{1}{r}=\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}[/tex], i.e. [tex]r=\frac{h_ah_bh_c}{h_ah_b+h_bh_c+h_ch_a}[/tex].

Solution:
From the right triangle formed by the altitude [tex]CH[/tex]: [tex]\angle BCH=90^\circ-\angle B[/tex].
The central angle is [tex]\angle AOC=2\angle B[/tex], and [tex]\triangle AOC[/tex] is isosceles, so [tex]\angle OCA=\frac{180^\circ-2\angle B}{2}=90^\circ-\angle B[/tex]. Hence [tex]\angle OCA=\angle BCH[/tex].
Since [tex]CL[/tex] bisects [tex]\angle ACB[/tex], [tex]\angle OCL=\angle ACL-\angle OCA=\angle BCL-\angle BCH=\angle HCL[/tex].

Solution:
[tex]\triangle AOD[/tex] is equilateral, so [tex]\angle AOD=60^\circ[/tex], and the inscribed angle on the same arc is [tex]\angle ABD=30^\circ[/tex].
The height is [tex]h=BD\sin 30^\circ=2[/tex], and the projection of the diagonal onto the longer base is the half-sum of the bases: [tex]\frac{AB+CD}{2}=BD\cos 30^\circ=2\sqrt{3}[/tex].
[tex]S=\frac{AB+CD}{2} \times h=4\sqrt{3}[/tex].

Solution:
[tex]\angle BAC=180^\circ-30^\circ-20^\circ=130^\circ[/tex], so [tex]P[/tex] lies on the extension of [tex]AC[/tex] beyond [tex]A[/tex] and [tex]\angle PAB=180^\circ-130^\circ=50^\circ[/tex].
The angle between the tangent and the chord [tex]BA[/tex] equals the inscribed angle on the same arc: [tex]\angle PBA=\angle ACB=30^\circ[/tex].
In [tex]\triangle PAB[/tex]: [tex]\angle APB=180^\circ-50^\circ-30^\circ=100^\circ[/tex].

Solution:
[tex]S=pr=336[/tex], so [tex]c=\frac{2S}{h_c}=42[/tex] and [tex]a+b=2p-c=54[/tex].
By Heron's formula [tex]S^2=p(p-a)(p-b)(p-c)[/tex]: [tex]336^2=48(48-a)(48-b) \times 6[/tex], i.e. [tex](48-a)(48-b)=392[/tex]; also [tex](48-a)+(48-b)=42[/tex]. So [tex]\{48-a;\ 48-b\}=\{14;\ 28\}[/tex] and [tex]a=34;\ b=20[/tex].
[tex]R=\frac{abc}{4S}=\frac{34 \times 20 \times 42}{4 \times 336}=\frac{85}{4}[/tex].

Solution:
For a cyclic quadrilateral Brahmagupta's formula holds: [tex]S=\sqrt{(p-a)(p-b)(p-c)(p-d)}[/tex], where [tex]p=\frac{2+3+4+7}{2}=8[/tex].
[tex]S=\sqrt{6 \times 5 \times 4 \times 1}=\sqrt{120}=2\sqrt{30}[/tex].

Solution:
[tex]O[/tex] lies on the bisectors of the angles at the endpoints of the leg [tex]AD[/tex], and these angles are supplementary, so [tex]\angle DAO+\angle ADO=\frac{\angle A+\angle D}{2}=90^\circ[/tex] and [tex]\angle AOD=90^\circ[/tex].
Then [tex]AD=\sqrt{36+64}=10[/tex]. The trapezoid is tangential and isosceles, so [tex]AB+CD=AD+BC=20[/tex] and [tex]P=40[/tex].

Solution:
The trapezoid is tangential: [tex]AB+CD=AD+BC[/tex], so [tex]BC=2[/tex].
Place [tex]A(0,0),\ B(4,0),\ D(x,h),\ C(x+1,h)[/tex]. Then [tex]x^2+h^2=9[/tex] and [tex](x-3)^2+h^2=4[/tex]; subtracting, [tex]x=\frac{7}{3};\ h^2=\frac{32}{9}[/tex].
[tex]BD^2=\left(4-\frac{7}{3}\right)^2+\frac{32}{9}=\frac{57}{9}[/tex], so [tex]BD=\frac{\sqrt{57}}{3}[/tex].

Solution:
Opposite angles of a cyclic quadrilateral are supplementary: [tex]\angle ABC+\angle CDA=180^\circ[/tex], so [tex]\angle ABC=\angle CDA=90^\circ[/tex].
In the right [tex]\triangle ABC[/tex]: [tex]\sin\angle BAC=\frac{BC}{AC}=\frac{1}{2}[/tex], so [tex]\angle BAC=30^\circ;\ \angle BCA=60^\circ[/tex]. The right [tex]\triangle ACD[/tex] is isosceles, so [tex]\angle DAC=\angle DCA=45^\circ[/tex].
Hence [tex]\angle A=75^\circ;\ \angle B=90^\circ;\ \angle C=105^\circ;\ \angle D=90^\circ[/tex].

Solution:
Let [tex]r[/tex] be the radius. The leg perpendicular to the bases is the height: [tex]h=2r[/tex]; the other leg is [tex]c=\frac{2r}{\sin\alpha}[/tex].
The trapezoid is tangential, so [tex]a+b=h+c=2r+\frac{2r}{\sin\alpha}[/tex], and [tex]S=\frac{a+b}{2} \times 2r=2r^2\left(1+\frac{1}{\sin\alpha}\right)=16[/tex].
Hence [tex]r^2=\frac{8\sin\alpha}{1+\sin\alpha}[/tex], and the area of the disk is [tex]\pi r^2=\frac{8\pi\sin\alpha}{1+\sin\alpha}[/tex].

Solution:
The distance from the center [tex]O[/tex] to the chord [tex]AB[/tex] is [tex]d_1=\sqrt{130-81}=7[/tex]. The other base [tex]CD[/tex] is a chord parallel to [tex]AB[/tex] at the distance of the height, so its distance from [tex]O[/tex] is either [tex]d_2=7-4=3[/tex] (both bases on the same side of [tex]O[/tex]) or [tex]d_2=7+4=11[/tex] (on opposite sides).
In the first case [tex]\frac{CD}{2}=\sqrt{130-9}=11[/tex], so [tex]CD=22[/tex] and [tex]S=\frac{18+22}{2} \times 4=80[/tex]; in the second case [tex]\frac{CD}{2}=\sqrt{130-121}=3[/tex], so [tex]CD=6[/tex] and [tex]S=\frac{18+6}{2} \times 4=48[/tex].

Solution:
Let the sides be [tex]7k;\ 15k;\ 20k[/tex]; then [tex]p=21k[/tex]. By Heron's formula [tex]S=\sqrt{21k \times 14k \times 6k \times k}=42k^2[/tex]; also [tex]S=pr=21k[/tex].
So [tex]42k^2=21k[/tex], i.e. [tex]k=\frac{1}{2}[/tex], and [tex]S=\frac{21}{2}[/tex].

Solution:
[tex]O[/tex] lies on the bisectors of the angles of the quadrilateral. From [tex]\triangle AOB[/tex]: [tex]\angle AOB=180^\circ-\frac{\angle A+\angle B}{2}[/tex]; from [tex]\triangle COD[/tex]: [tex]\angle COD=180^\circ-\frac{\angle C+\angle D}{2}[/tex].
The angles of the quadrilateral add up to a full angle, so [tex]\angle AOB+\angle COD=360^\circ-\frac{\angle A+\angle B+\angle C+\angle D}{2}=360^\circ-180^\circ=180^\circ[/tex].

Solution:
The angle between the tangent [tex]t[/tex] and the chord [tex]CA[/tex] equals the inscribed angle on the same arc: [tex]\angle(t;CA)=\angle ABC[/tex].
Since [tex]\angle AA_1B=\angle AB_1B=90^\circ[/tex], the quadrilateral [tex]ABA_1B_1[/tex] is cyclic, so its exterior angle [tex]\angle CB_1A_1=\angle ABC[/tex].
So [tex]\angle CB_1A_1=\angle(t;CA)[/tex]; these are alternate angles at the transversal [tex]CA[/tex], hence [tex]A_1B_1 \parallel t[/tex].

Solution:
As proved in the problem [tex]AH=2OM[/tex] on this page, [tex]AH=2R\cos A[/tex]; by the Law of Sines [tex]BC=2R\sin A[/tex].
So [tex]\cos A=\sin A[/tex], i.e. [tex]\tanA=1[/tex], and since the angle is acute, [tex]\angle BAC=45^\circ[/tex].

Solution:
As proved in the problem [tex]AH=2OM[/tex] on this page, [tex]\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}[/tex]. The centroid satisfies [tex]\overrightarrow{OG}=\frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}[/tex].
So [tex]\overrightarrow{OH}=3\overrightarrow{OG}[/tex], hence [tex]O[/tex], [tex]G[/tex], [tex]H[/tex] are collinear, and [tex]\overrightarrow{GH}=\overrightarrow{OH}-\overrightarrow{OG}=2\overrightarrow{OG}[/tex], i.e. [tex]GH=2OG[/tex].

Solution:
Let [tex]O[/tex] be the center (the midpoint of [tex]AC[/tex]) and [tex]K[/tex] its projection on [tex]BD[/tex]. Since [tex]OK \perp BD[/tex] and [tex]O[/tex] is the center, [tex]BK=KD[/tex].
Also [tex]AM \parallel OK \parallel CN[/tex], and [tex]O[/tex] is the midpoint of [tex]AC[/tex], so by Thales' theorem [tex]MK=KN[/tex].
Hence [tex]BM=DN[/tex].

Solution:
By the property proved earlier on this page (the circles circumscribed about [tex]\triangle ABC[/tex] and [tex]\triangle ABH[/tex] have equal radii), [tex]R_{ABH}=R[/tex]. So [tex]OA=OB=O'A=O'B=R[/tex], and [tex]AOBO'[/tex] is a rhombus (the points [tex]O[/tex] and [tex]O'[/tex] are different, because [tex]H[/tex] does not lie on the circumscribed circle of [tex]\triangle ABC[/tex]).
The diagonals of a rhombus are perpendicular and bisect each other, so [tex]AB[/tex] is the perpendicular bisector of [tex]OO'[/tex].

Solution:
The angles subtending the diameter are right: [tex]\angle CAD=\angle CBD=90^\circ[/tex], so [tex]DB \perp BC[/tex]; also [tex]AH \perp BC[/tex]. Hence [tex]AH \parallel DB[/tex], and in the same way [tex]BH \parallel DA[/tex].
So [tex]AHBD[/tex] is a parallelogram, and its diagonals [tex]AB[/tex] and [tex]HD[/tex] bisect each other: the midpoint [tex]M[/tex] of [tex]AB[/tex] is also the midpoint of [tex]HD[/tex].

Solution:
Let [tex]a=BC[/tex], [tex]b=CA[/tex], [tex]c=AB[/tex]. The tangent segments give [tex]AK=p-a[/tex] and [tex]KB=p-b[/tex], so [tex]AK \times KB=\frac{(c+b-a)(c+a-b)}{4}=\frac{c^2-(a-b)^2}{4}[/tex].
Since [tex]c^2=a^2+b^2[/tex], [tex]\frac{c^2-(a-b)^2}{4}=\frac{2ab}{4}=\frac{ab}{2}=S[/tex].

Solution:
Let [tex]\angle C=90^\circ[/tex]. Then [tex]c=2R[/tex], [tex]r=\frac{a+b-c}{2}[/tex] and [tex]a=c\sin A;\ b=c\cos A[/tex], so [tex]\frac{r}{R}=\sin A+\cos A-1[/tex].
Since [tex]\sin A+\cos A=\sqrt{2}\sin(A+45^\circ) \le \sqrt{2}[/tex], [tex]\frac{r}{R} \le \sqrt{2}-1[/tex]; equality holds for [tex]A=45^\circ[/tex].