Inscribed and Circumscribed Circles, Inscribed Angles: Problems with Solutions

Solution:
[tex]\angle AOB[/tex] is a central angle and [tex]\angle ACB=70^\circ[/tex] is an inscribed angle subtending the same arc [tex]AB[/tex].
A central angle is twice the inscribed angle on the same arc, so [tex]\angle AOB=2\angle ACB=140^\circ[/tex].

Solution:
The angle of [tex]120^\circ[/tex] can only be the vertex angle: [tex]AC=BC=6[/tex], [tex]\angle ACB=120^\circ[/tex]. Then [tex]\angle A=\angle B=30^\circ[/tex].
By the Law of Sines [tex]BC=2R\sin\angle A[/tex], i.e. [tex]6=2R \times \frac{1}{2}[/tex], so [tex]R=6[/tex].

Solution:
A circle can be inscribed in a quadrilateral exactly when the sums of opposite sides are equal: [tex]AB+CD=AD+BC[/tex], so [tex]AD=BC=\frac{32+18}{2}=25[/tex].
The projection of a leg onto the longer base is [tex]\frac{32-18}{2}=7[/tex], so the height is [tex]h=\sqrt{25^2-7^2}=24[/tex].
The height of the trapezoid equals the diameter of the circle, so [tex]r=\frac{h}{2}=12[/tex].

Solution:
The two tangent segments drawn from a point to a circle are equal: [tex]AP=AM=6;\ BM=BN=8;\ CN=CP=9[/tex].
Then [tex]AB=6+8=14;\ BC=8+9=17;\ AC=9+6=15[/tex].

Solution:
In a cyclic quadrilateral opposite angles add up to [tex]180^\circ[/tex]: [tex]\angle A+\angle C=180^\circ[/tex].
[tex]3k+7k=180^\circ[/tex], so [tex]k=18^\circ[/tex].
Then [tex]\angle A=54^\circ;\ \angle B=72^\circ;\ \angle C=126^\circ[/tex] and [tex]\angle D=180^\circ-\angle B=108^\circ[/tex].

Solution:
The semiperimeter is [tex]p=\frac{13+14+15}{2}=21[/tex].
By Heron's formula [tex]S=\sqrt{21 \times 8 \times 7 \times 6}=84[/tex].
[tex]r=\frac{S}{p}=\frac{84}{21}=4[/tex].

Solution:
Let [tex]AC=3[/tex], [tex]BC=4[/tex], so [tex]AB=5[/tex]. The circumcenter [tex]O[/tex] is the midpoint of the hypotenuse [tex]AB[/tex] and [tex]R=\frac{5}{2}[/tex].
The inradius of a right triangle is [tex]r=\frac{3+4-5}{2}=1[/tex]. Let the incircle (center [tex]I[/tex]) touch [tex]AB[/tex] at [tex]T[/tex]; then [tex]AT=p-BC=6-4=2[/tex].
Hence [tex]OT=AO-AT=\frac{5}{2}-2=\frac{1}{2}[/tex], and since [tex]IT \perp AB[/tex]: [tex]OI^2=OT^2+IT^2=\frac{1}{4}+1=\frac{5}{4}[/tex], i.e. [tex]OI=\frac{\sqrt{5}}{2}[/tex].

Solution:
The diagonal of the square is a diameter of the circle: [tex]d=a\sqrt{2}=\sqrt{2} \times \sqrt{2}=2[/tex].
So [tex]R=\frac{d}{2}=1[/tex] and [tex]C=2\pi R=2\pi[/tex].

Solution:
The inscribed angle [tex]\angle AMB[/tex] subtends the arc [tex]AB[/tex] that does not contain [tex]M[/tex]; its measure is [tex]360^\circ-108^\circ=252^\circ[/tex].
An inscribed angle is half of the arc it subtends, so [tex]\angle AMB=\frac{252^\circ}{2}=126^\circ[/tex].

Solution:
Let [tex]AB=6[/tex], [tex]AC=BC[/tex], and let the incircle touch [tex]AC[/tex] at [tex]M[/tex], so [tex]CM=7[/tex].
The incircle touches the base at its midpoint, and the tangent segments from [tex]A[/tex] are equal, so [tex]AM=\frac{AB}{2}=3[/tex].
Hence [tex]AC=CM+AM=7+3=10[/tex].

Solution:
A trapezoid inscribed in a circle is isosceles.
The projection of each leg onto the longer base is [tex]\frac{40-14}{2}=13[/tex].
By the Pythagorean theorem [tex]AD=BC=\sqrt{13^2+39^2}=\sqrt{1690}=13\sqrt{10}[/tex].

Solution:
The three equal arcs make up a semicircle, so [tex]\overset{\frown}{AD}=\overset{\frown}{DC}=\overset{\frown}{CB}=60^\circ[/tex] and [tex]R=1[/tex].
A chord subtending an arc of [tex]60^\circ[/tex] equals the radius (it is a side of the inscribed regular hexagon), so [tex]AD=DC=CB=R=1[/tex].
[tex]\angle A[/tex] subtends [tex]\overset{\frown}{DCB}=120^\circ[/tex], so [tex]\angle A=\frac{120^\circ}{2}=60^\circ[/tex]; similarly [tex]\angle B=60^\circ[/tex], and since the quadrilateral is cyclic, [tex]\angle C=\angle D=120^\circ[/tex].

Solution:
A regular hexagon consists of six equilateral triangles with side [tex]6[/tex]; the inradius is the altitude of one of them: [tex]r=\frac{a\sqrt{3}}{2}=\frac{6\sqrt{3}}{2}=3\sqrt{3}[/tex].

Solution:
The height of the rhombus equals the diameter of the inscribed circle: [tex]h=2r=14[/tex].
[tex]\angle A=180^\circ-150^\circ=30^\circ[/tex], so the side is [tex]a=\frac{h}{\sin 30^\circ}=28[/tex].
[tex]S=a \times h=28 \times 14=392[/tex].

Solution:
In a tangential quadrilateral the sums of opposite sides are equal: [tex]a+b=c+d[/tex], so the sum of the bases is half the perimeter: [tex]a+b=\frac{32}{2}=16[/tex].
The midsegment is [tex]m=\frac{a+b}{2}=8[/tex].

Solution:
For a right triangle with legs [tex]a[/tex], [tex]b[/tex] and hypotenuse [tex]c[/tex], [tex]r=\frac{a+b-c}{2}[/tex], because the tangent segments from the vertex of the right angle are equal to the radius.
Hence [tex]a+b=2r+c=4+13=17[/tex].

Solution:
[tex]AJ[/tex] and [tex]BJ[/tex] are angle bisectors, so [tex]\angle AJB=180^\circ-\frac{\angle A+\angle B}{2}=90^\circ+\frac{\angle C}{2}[/tex].
Then [tex]\frac{\angle C}{2}=25^\circ[/tex] and [tex]\angle ACB=50^\circ[/tex].

Solution:
[tex]\triangle AOB[/tex] has three equal sides, so it is equilateral and [tex]\angle AOB=60^\circ[/tex].
The inscribed angle is half of the central angle on the same arc: [tex]\angle ACB=\frac{60^\circ}{2}=30^\circ[/tex].

Solution:
[tex]OA=OB=OC[/tex], so [tex]\triangle AOC[/tex] and [tex]\triangle BOC[/tex] are isosceles: [tex]\angle OAC=\angle OCA[/tex] and [tex]\angle OBC=\angle OCB[/tex].
The sum of the angles of [tex]\triangle ABC[/tex] gives [tex]2\angle OCA+2\angle OCB=180^\circ[/tex].
Hence [tex]\angle ACB=\angle OCA+\angle OCB=90^\circ[/tex].

Solution:
The radii of the excircles are [tex]r_a=\frac{S}{p-a}[/tex], [tex]r_b=\frac{S}{p-b}[/tex] and [tex]r_c=\frac{S}{p-c}[/tex], where [tex]S[/tex] is the area and [tex]p[/tex] the semiperimeter.
In an equilateral triangle [tex]a=b=c[/tex], so [tex]p-a=p-b=p-c[/tex] and therefore [tex]r_a=r_b=r_c[/tex].

Solution:
Let the side be [tex]a[/tex]. Then [tex]r=\frac{S}{p}[/tex] and [tex]r_a=\frac{S}{p-a}[/tex].
[tex]p=\frac{3a}{2}[/tex], so [tex]p-a=\frac{a}{2}=\frac{p}{3}[/tex].
Hence [tex]r_a=3r=12[/tex].

Solution:
The circle is inscribed in the square, so its diameter equals the side: [tex]r=\frac{14}{2}=7[/tex].
[tex]S=\pi r^2=49\pi[/tex].

Solution:
In an equilateral triangle the centers of both circles coincide with the centroid, which divides each altitude in the ratio [tex]2:1[/tex] from the vertex.
So [tex]R=\frac{2}{3}h=10[/tex] and [tex]r=\frac{1}{3}h=5[/tex].

Solution:
An inscribed angle subtending a diameter is right: [tex]\angle CAD=90^\circ[/tex].
[tex]\angle ADC=\angle ABC=60^\circ[/tex], since both inscribed angles subtend the arc [tex]AC[/tex].
[tex]\angle ACD=180^\circ-90^\circ-60^\circ=30^\circ[/tex].

Solution:
The hypotenuse of a right triangle inscribed in a circle is a diameter: [tex]c=13[/tex].
The other leg is [tex]b=\sqrt{13^2-5^2}=12[/tex], so [tex]S=\frac{5 \times 12}{2}=30[/tex].

Solution:
The altitude bisects the base, so the leg is [tex]b=\sqrt{8^2+6^2}=10[/tex].
[tex]S=\frac{16 \times 6}{2}=48[/tex], [tex]p=\frac{16+10+10}{2}=18[/tex].
[tex]r=\frac{S}{p}=\frac{48}{18}=\frac{8}{3}[/tex].

Solution:
[tex]2\pi R=\pi R^2[/tex], so [tex]R=2[/tex].
The diagonal of the square is a diameter, so [tex]a=R\sqrt{2}=2\sqrt{2}[/tex].

Solution:
A trapezoid inscribed in a circle is isosceles: [tex]AD=BC[/tex].
The projection of a leg onto the longer base is [tex]\frac{AB-CD}{2}=1[/tex], so [tex]AD=\frac{1}{\cos 60^\circ}=2[/tex].

Solution:
Let [tex]ABCD[/tex] be the rhombus. Its opposite angles are equal: [tex]\angle A=\angle C[/tex].
It is cyclic, so [tex]\angle A+\angle C=180^\circ[/tex]. Hence [tex]\angle A=\angle C=90^\circ[/tex], and similarly [tex]\angle B=\angle D=90^\circ[/tex].
A rhombus with right angles is a square.

Solution:
The angle subtending the diameter is right: [tex]\angle ADB=90^\circ[/tex], so [tex]\angle DAB=90^\circ-40^\circ=50^\circ[/tex].
A trapezoid inscribed in a circle is isosceles, so [tex]\angle A=\angle B=50^\circ[/tex] and [tex]\angle C=\angle D=130^\circ[/tex].

Solution:
The inscribed angle is half of the central angle on the same arc: [tex]\angle ABC=\frac{60^\circ}{2}=30^\circ[/tex].
[tex]AJ[/tex] and [tex]CJ[/tex] are angle bisectors, so [tex]\angle AJC=180^\circ-\frac{\angle A+\angle C}{2}=90^\circ+\frac{\angle ABC}{2}=105^\circ[/tex].

Solution:
Let [tex]O[/tex] be the center and [tex]P[/tex] the intersection of the tangents. The central angle is [tex]\angle AOB=47^\circ 20'[/tex], and the radii are perpendicular to the tangents: [tex]\angle OAP=\angle OBP=90^\circ[/tex].
In the quadrilateral [tex]OAPB[/tex]: [tex]\angle APB=180^\circ-47^\circ 20'=132^\circ 40'[/tex].

Solution:
The trapezoid is tangential, so [tex]a+b=2 \times 14=28[/tex]. From [tex]5k+2k=28[/tex]: [tex]k=4[/tex], so [tex]a=20[/tex] and [tex]b=8[/tex].
The projection of a leg onto the longer base is [tex]\frac{20-8}{2}=6[/tex], so [tex]h=\sqrt{14^2-6^2}=\sqrt{160}=4\sqrt{10}[/tex].
[tex]r=\frac{h}{2}=2\sqrt{10}[/tex].

Solution:
Let the side be [tex]a[/tex]. The radius of the circumscribed circle is half the diagonal: [tex]R=\frac{a\sqrt{2}}{2}[/tex]; the radius of the inscribed circle is half the side: [tex]r=\frac{a}{2}[/tex].
[tex]\frac{S_1}{S_2}=\frac{\pi R^2}{\pi r^2}=\left(\frac{R}{r}\right)^2=(\sqrt{2})^2=2[/tex].

Solution:
[tex]9x=360^\circ[/tex], so [tex]\overset{\frown}{AB}=80^\circ;\ \overset{\frown}{CA}=160^\circ[/tex].
The angle between a tangent and a chord equals half of the arc the chord cuts off, so the tangent makes [tex]\frac{80^\circ}{2}=40^\circ[/tex] with [tex]AB[/tex] and [tex]\frac{160^\circ}{2}=80^\circ[/tex] with [tex]AC[/tex].

Solution:
Let [tex]r[/tex] be the inradius. The tangent segments from the right-angle vertex [tex]C[/tex] equal [tex]r[/tex], so the hypotenuse is [tex]AB=4+6=10[/tex] and the legs are [tex]AC=r+4;\ BC=r+6[/tex].
By the Pythagorean theorem [tex](r+4)^2+(r+6)^2=10^2[/tex], i.e. [tex]r^2+10r-24=0[/tex], so [tex]r=2[/tex] and [tex]AC=6;\ BC=8[/tex].
[tex]S=\frac{6 \times 8}{2}=24[/tex].

Solution:
The tangent segments from each vertex are equal, so in a quadrilateral circumscribed about a circle the sums of opposite sides are equal: [tex]AB+CD=BC+DA[/tex].
Hence [tex]DA=5+9-7=7[/tex].

Solution:
Let the legs be [tex]c[/tex]. The trapezoid is tangential, so [tex]a+b=2c[/tex], i.e. [tex]c=\frac{a+b}{2}[/tex].
The projection of a leg onto the longer base is half the difference of the bases, so the height [tex]h[/tex] satisfies [tex]h^2=c^2-\left(\frac{a-b}{2}\right)^2=\frac{(a+b)^2-(a-b)^2}{4}=ab[/tex].
The height equals the diameter: [tex]h=2r[/tex], hence [tex](2r)^2=ab[/tex], i.e. [tex]2r=\sqrt{ab}[/tex].

Solution:
Since [tex]\angle ADC=\angle BDC=90^\circ[/tex], the hypotenuses of [tex]\triangle ACD[/tex] and [tex]\triangle BCD[/tex] are [tex]AC[/tex] and [tex]BC[/tex], so the centers [tex]O_1[/tex] and [tex]O_2[/tex] are the midpoints of [tex]AC[/tex] and [tex]BC[/tex].
Then [tex]O_1O_2[/tex] is a midsegment of [tex]\triangle ABC[/tex], so [tex]O_1O_2 \parallel AB[/tex], and it bisects every segment joining [tex]C[/tex] with a point of [tex]AB[/tex], in particular [tex]CD[/tex].

Solution:
[tex]O[/tex] lies on the angle bisectors, so [tex]\angle AOB=90^\circ+\frac{\angle ACB}{2}=150^\circ[/tex].
[tex]S=\frac{1}{2} \times 4\sqrt{2} \times 3 \times \sin 150^\circ=3\sqrt{2}[/tex].

Solution:
[tex]I[/tex] lies on the bisector of the angle at [tex]C[/tex] and on [tex]CH[/tex], so the altitude [tex]CH[/tex] is also the angle bisector: [tex]\angle ACI=\angle BCI[/tex].
In the right triangles [tex]\triangle ACH[/tex] and [tex]\triangle BCH[/tex]: [tex]\angle CAH=90^\circ-\angle ACH=90^\circ-\angle BCH=\angle CBH[/tex].
The base angles are equal, so [tex]AC=BC[/tex].

Solution:
Let the trapezoid be [tex]ABCD[/tex] with diagonal [tex]BD[/tex]. The circle circumscribed about the trapezoid is also circumscribed about [tex]\triangle ABD[/tex], in which [tex]\angle A=45^\circ[/tex] is opposite [tex]BD[/tex].
By the Law of Sines [tex]R=\frac{BD}{2\sin 45^\circ}=\frac{10}{\sqrt{2}}=5\sqrt{2}[/tex].

Solution:
Let [tex]\angle A=\angle B=\alpha[/tex]. By the Law of Sines [tex]BC=2R\sin\alpha[/tex], so [tex]\sin\alpha=\frac{1}{2}[/tex].
The base angles are acute, so [tex]\alpha=30^\circ[/tex] and [tex]\angle C=180^\circ-2 \times 30^\circ=120^\circ[/tex].

Solution:
Corresponding angles of congruent triangles are equal: [tex]\angle ACB=\angle BDA[/tex].
So the segment [tex]AB[/tex] is seen from [tex]C[/tex] and [tex]D[/tex], which lie on the same side of [tex]AB[/tex], at equal angles. Hence [tex]C[/tex] and [tex]D[/tex] lie on the same arc over [tex]AB[/tex], i.e. [tex]ABCD[/tex] is cyclic.

Solution:
The incircle touches the base [tex]AB[/tex] at its midpoint [tex]M[/tex], so [tex]AM=2\sqrt{3}[/tex], and its center [tex]I[/tex] lies on the bisector of the angle at [tex]A[/tex]: [tex]\tan\angle IAM=\frac{IM}{AM}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}[/tex], so [tex]\angle IAM=30^\circ[/tex] and [tex]\angle BAC=60^\circ[/tex].
The base angles are equal, so [tex]\angle ABC=\angle BAC=60^\circ[/tex] and [tex]\angle ACB=60^\circ[/tex]: the triangle is equilateral.

Solution:
Extend the altitude [tex]CH[/tex] to the diameter [tex]CD[/tex]; then [tex]\angle CAD=90^\circ[/tex] (it subtends a diameter) and [tex]CD=2R=25[/tex].
In the right [tex]\triangle ACD[/tex] the leg is the geometric mean of its projection and the hypotenuse: [tex]AC^2=CH \times CD=16 \times 25=400[/tex], so [tex]AC=20[/tex].
Then [tex]AH=\sqrt{400-256}=12[/tex] and [tex]AB=24[/tex].

Solution:
Place [tex]C(0,0),\ A(12,0),\ B(0,9)[/tex]; then [tex]c=15[/tex] and [tex]r=\frac{9+12-15}{2}=3[/tex], so [tex]I(3,3)[/tex].
The centroid is [tex]G\left(\frac{0+12+0}{3},\frac{0+0+9}{3}\right)=G(4,3)[/tex].
Both points have the same ordinate, so [tex]GI=4-3=1[/tex].

Solution:
Equal chords subtend equal arcs: [tex]\overset{\frown}{AD}=\overset{\frown}{BC}[/tex].
The inscribed angles on these arcs are equal: [tex]\angle ABD=\angle BDC[/tex]. They are alternate angles at the transversal [tex]BD[/tex], so [tex]AB \parallel CD[/tex].

Solution:
The diagonals of a rhombus are perpendicular and bisect each other: [tex]AC \perp BD[/tex], [tex]BP=PD[/tex]. So [tex]M[/tex] lies on the perpendicular bisector of [tex]BD[/tex], hence [tex]MB=MD[/tex]; also [tex]AB=AD[/tex].
Then [tex]AB+MD=AD+MB[/tex], and a convex quadrilateral with equal sums of opposite sides has an inscribed circle.

Solution:
Let [tex]K[/tex] be the point of tangency on [tex]AB[/tex]; then [tex]OK=r[/tex] and [tex]\triangle AOK[/tex] is right. [tex]O[/tex] lies on the bisector of the angle at [tex]A[/tex], so [tex]\angle OAK=\frac{\alpha}{2}[/tex] and [tex]OA=\frac{OK}{\sin\frac{\alpha}{2}}=\frac{r}{\sin\frac{\alpha}{2}}[/tex].
In the same way [tex]OB=\frac{r}{\sin\frac{\beta}{2}};\ OC=\frac{r}{\sin\frac{\gamma}{2}}[/tex].

Solution:
The inscribed angle is half of the central angle on the same arc: [tex]\angle BAC=\frac{\angle BOC}{2}=45^\circ[/tex], so [tex]\angle ABC=\angle BAC=45^\circ[/tex] and [tex]\angle ACB=90^\circ[/tex].
Then [tex]AB=2R[/tex] is a diameter, and the altitude to it is [tex]CH=R[/tex] (it is the radius [tex]OC[/tex]).
[tex]S=\frac{AB \times CH}{2}=\frac{2R \times R}{2}=R^2[/tex].

Solution:
Join [tex]O[/tex] with the vertex of the right angle; the radii to the points of tangency are the altitudes of [tex]\triangle AOC[/tex] and [tex]\triangle BOC[/tex] to the legs. The area of the triangle is [tex]S=\frac{5 \times 12}{2}=30[/tex], and also [tex]S=\frac{r \times 12}{2}+\frac{r \times 5}{2}=\frac{17r}{2}[/tex].
So [tex]\frac{17r}{2}=30[/tex] and [tex]r=\frac{60}{17}[/tex].

Solution:
The incircle touches the base at its midpoint [tex]M[/tex], so [tex]AM=a[/tex], and its center [tex]I[/tex] lies on the bisector of the angle at [tex]A[/tex]: [tex]\angle IAM=\frac{\alpha}{2}[/tex]. From the right [tex]\triangle AIM[/tex]: [tex]r=a\tan\frac{\alpha}{2}[/tex].
[tex]\angle ACB=180^\circ-2\alpha[/tex], and by the Law of Sines [tex]R=\frac{AB}{2\sin\angle ACB}=\frac{2a}{2\sin 2\alpha}=\frac{a}{\sin 2\alpha}[/tex].

Solution:
The interior and exterior angles at [tex]C[/tex] are supplementary, so their bisectors [tex]CP[/tex] and [tex]CQ[/tex] are perpendicular: [tex]\angle PCQ=90^\circ[/tex].
An inscribed right angle subtends a diameter, so [tex]PQ[/tex] is a diameter.

Solution:
Extend [tex]AP[/tex] beyond [tex]P[/tex] to meet the circle at [tex]Q[/tex]. The angle subtending the diameter is right: [tex]\angle AQB=90^\circ[/tex].
[tex]\angle APB[/tex] is an exterior angle of [tex]\triangle PQB[/tex], so [tex]\angle APB=\angle AQB+\angle QBP>90^\circ[/tex].

Solution:
On the one hand [tex]S=\frac{ch}{2}[/tex]; on the other hand, with semiperimeter [tex]p=\frac{2b+c}{2}[/tex], [tex]S=pr[/tex].
So [tex]\frac{ch}{2}=\frac{2b+c}{2}r[/tex], which gives [tex]r=\frac{ch}{2b+c}[/tex].

Solution:
The diagonals of a rhombus bisect its angles, so [tex]O[/tex] lies on the bisectors of all four angles.
A point on an angle bisector is equidistant from the sides of the angle, so the distances from [tex]O[/tex] to all four sides are equal to its distance to [tex]AB[/tex], i.e. to [tex]r[/tex]. Hence [tex]k[/tex] touches [tex]BC;\ CD;\ DA[/tex] as well.