AMC 10 Step by Step
Printable problems

Circles

Chords, tangents, inscribed angles, power of a point, arc length, sectors, inscribed/circumscribed circles

What you need to know

  • Inscribed Angle Theorem. An inscribed angle (corner on the circle) is half the arc it cuts off. So an inscribed angle whose sides pass through the two ends of a diameter is a right angle. Example: an inscribed angle on a 140∘140^\circ arc is 70∘70^\circ.
  • Tangents. A tangent line is perpendicular to the radius at the point where it touches. Two Tangent Theorem: the two tangent segments from a point outside the circle are equal. Example: from a point 1010 away from the center of a circle with radius 66, both tangents are 100−36=8\sqrt{100-36} = 8 long.
  • Power of a Point. If chords ABAB and CDCD cross at PP, then PA⋅PB=PC⋅PDPA\cdot PB = PC\cdot PD. From a point PP outside the circle, with a tangent PTPT and a line (a secant) meeting the circle at AA and then BB: PT2=PA⋅PBPT^2 = PA\cdot PB. Example: chords with PA=4PA = 4, PB=6PB = 6, PC=3PC = 3 give PD=8PD = 8. A secant with PA=4PA = 4, PB=9PB = 9 gives PT=6PT = 6.
  • Arcs and sectors. A central angle of θ\theta degrees cuts off θ360∘\frac{\theta}{360^\circ} of the circle: the arc length is θ360∘⋅2πr\frac{\theta}{360^\circ}\cdot 2\pi r, and the sector (pizza slice) has area θ360∘⋅πr2\frac{\theta}{360^\circ}\cdot \pi r^2. Example: with r=6r = 6 and θ=60∘\theta = 60^\circ, the arc is 2π2\pi long and the sector's area is 6π6\pi.

How AMC 10 tests it

  • Shaded regions (problems 3–12): circles in squares, or semicircles on the sides of a triangle. Subtract sectors and triangles.
  • Tangent lines from a point, or one line tangent to two circles. Radii to the points of tangency give a right triangle or a trapezoid with two right angles.
  • Problems 17–23: Power of a Point for a length, or a rope wrapping around a polygon, giving a sum of sectors.

Standard approaches

  1. Draw the radius to every point of tangency and every chord end. Look for right angles, and for isosceles triangles (radii are equal).
  2. Turn inscribed angles into arcs, add the arcs, and turn back.
  3. When two chords, or a tangent and a secant, pass through one point, write Power of a Point.

Worked example

Point PP lies outside a circle with center OO and radius 66. The two tangent segments from PP touch the circle at AA and BB, and ∠APB=60∘\angle APB = 60^\circ. What is the area of the region bounded by PA‾\overline{PA}, PB‾\overline{PB}, and the minor arc ABAB?

(A) 183−6π18\sqrt3 - 6\pi (B) 363−6π36\sqrt3 - 6\pi (C) 363−12π36\sqrt3 - 12\pi (D) 36−12π36 - 12\pi (E) 723−12π72\sqrt3 - 12\pi

Solution. Step 1. Each tangent is perpendicular to its radius, so ∠OAP=∠OBP=90∘\angle OAP = \angle OBP = 90^\circ.

Step 2. Right triangles OAPOAP and OBPOBP share the hypotenuse OPOP and have OA=OBOA = OB, so they are congruent, and ∠APO=12⋅60∘=30∘\angle APO = \frac12\cdot60^\circ = 30^\circ. Triangle OAPOAP is a 3030-6060-9090 triangle with short leg OA=6OA = 6, so PA=63PA = 6\sqrt3 and ∠AOP=60∘\angle AOP = 60^\circ.

Step 3. The kite OAPBOAPB is two copies of triangle OAPOAP, so its area is 2⋅12⋅6⋅63=3632\cdot\frac12\cdot 6\cdot 6\sqrt3 = 36\sqrt3.

Step 4. The central angle is ∠AOB=120∘\angle AOB = 120^\circ, so sector OABOAB has area 120360⋅36π=12π\frac{120}{360}\cdot 36\pi = 12\pi.

Step 5. The region is the kite minus the sector: 363−12π36\sqrt3 - 12\pi.

The answer is (C) 363−12π\boxed{\textbf{(C)}\ 36\sqrt3 - 12\pi}.

Pitfalls

  • Putting the diameter into πr2\pi r^2 instead of the radius. The area comes out 44 times too big, and the answer choices often differ by that factor of 44.
  • Finding a sector's area with the inscribed angle instead of the central angle (which is twice as big).
  • In Power of a Point, using PA⋅ABPA\cdot AB instead of PA⋅PBPA\cdot PB. Both lengths start at PP. Example: with PA=4PA = 4 and AB=5AB = 5, the product is 4⋅9=364\cdot 9 = 36, not 2020.

Traps that recur

  • Assuming the inner corners lie on a circle of radius 3 (the table radius minus the mat width), rather than at distance x from the center. (2008 AMC 10A #25)
  • Assuming the inner circle is the incircle of the triangle of centres, whose inradius is 1, or reading the picture's apparent symmetry as meaning the inner centre is equidistant from the three outer centres. (2025 AMC 10A #22)
  • Losing a minus sign when writing the diameter, giving (sqrt 2 - 1)(sqrt 3 + 1) and the sum 5 in (B). (2025 AMC 10B #20)
  • Finding only the small pipe wedged between the two and answering 1/9, forgetting that a pipe on the far side of the small one is also in contact with both. (2024 AMC 10B #21)

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.