AMC 10 Daily Practice - Tangency

Complete problem set with solutions and individual problem pages

Problem 1 Medium

As shown in the figure, AC is a tangent to circle \odot O, and B is the point of tangency. Connect OA and OC. If m\angle A = 30^\circ, AB = 2\sqrt{3}, and BC = 3, then what is the length of OC?

 

 

  • A.

    3

  • B.

    2\sqrt{3}

  • C.

    \sqrt{13}

  • D.

    6

  • E.

    4

Answer:C

Connect OB

\because AC is tangent to circle \odot O, point B is the tangent point，

\therefore \overline{OB}\bot \overline{AC},

\because m\angle A=30{}^\circ, AB=2\sqrt{3},

\therefore In \text{Rt}\triangle OAB, OB=2\sqrt{3}\times \frac{\sqrt{3}}{3}=2,

\because BC=3,

\therefore In \text{Rt}\triangle OBC, OC=\sqrt{O{{B}^{2}}+B{{C}^{2}}}=\sqrt{13},

choose \text{C}.

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Problem 2 Easy

As shown in the figure, the diameter of each circle is 8\text{cm}. Therefore, the value of "?" is            \text{cm}.

  • A.

    4 ( 1 + 2 \sqrt { 3 } )

  • B.

    4 ( 2 + \sqrt { 3 } )

  • C.

    24

  • D.

    8 ( 1 + \sqrt { 3 } )

  • E.

    8 ( 2 + \sqrt { 3 } )

Answer:D

First, form a equilateral triangle ：

​​

Diameter is 8 cm, thus length of triangle is 16 cm.

Then divide the triangle through the top vertex.

Apply the Pythagorean Theorem c^2 = a^2 + b^2, to calculate the height of the triangle,

256 = 64 + h^2, h = 8\sqrt{3}

Add the miss part from the top and bottom,

? = 4(2) + 8\sqrt{3}

? = 8 + 8\sqrt{3} = 8(1 + \sqrt{3})

Thus, the answer is D.

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Problem 3 Easy

Circle C_1 and C_2 each have radius 1, and the distance between their centers is \frac{1}{2}. Circle C_3 is the largest circle internally tangent to both C_1 and C_2. Circle C_4 is internally tangent to both C_1 and C_2 and externally tangent to C_3. What is the radius of C_4? (2023 AMC 10A Problems, Quetsion #22)

  • A.

    \frac{1}{14}

  • B.

    \frac{1}{12}

  • C.

    \frac{1}{10}

  • D.

    \frac{3}{28}

  • E.

    \frac{1}{9}

Answer:D

Let O be the center of the midpoint of the line segment connecting both the centers, say A and B.

 

Let the point of tangency with the inscribed circle and the right larger circles be T.

 

Then OT = BO + BT = BO + AT - \frac{1}{2} = \frac{1}{4} + 1 - \frac{1}{2} = \frac{3}{4}.

 

Since C_4 is internally tangent to C_1, center of C_4, C_1 and their tangent point must be on the same line.

 

Now, if we connect centers of C_4, C_3 and C_1/C_2, we get a right angled triangle.

 

Let the radius of C_4 equal r. With the pythagorean theorem on our triangle, we have

 

\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2

 

Solving this equation gives us

 

r = \boxed{\textbf{(D) } \frac{3}{28}}

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Related Paper
AMC 10 Daily Practice - Modular Arithmetic AMC 10 Daily Practice - Polygons AMC 10 Daily Practice - Similarity AMC 10 Daily Practice - Polynomial Functions
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