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2007 AMC 10B Problem 23

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2007 AMC 10B Problem 23AMC 10 Step by Step
Name:                             Date: September 13, 2026
Solid Geometry · difficulty 4 of 5 · about 4 min · position P21-25

A pyramid with a square base is cut by a plane that is parallel to its base and 22 units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?

  1. A)

    22

  2. B)

    2+22+\sqrt{2}

  3. C)

    1+221+2\sqrt{2}

  4. D)

    44

  5. E)

    4+224+2\sqrt{2}

Answer

E

Key insight

The top pyramid is a scaled copy; half the surface area means scale factor 1/sqrt 2, so its height is H/sqrt 2 and H - H/sqrt 2 = 2.

Solution

A plane parallel to the base cuts off a pyramid similar to the original. If the original has altitude HH, the small pyramid has altitude H−2H - 2, so the similarity ratio is k=H−2Hk = \frac{H - 2}{H}.

Surface area (base plus lateral faces, all scaled the same way) is proportional to the square of the ratio:
k2=12⟹k=12.k^2 = \frac12 \quad\Longrightarrow\quad k = \frac{1}{\sqrt2}.
Therefore
H−2H=12⟹H−2=H2⟹H(1−12)=2.\frac{H - 2}{H} = \frac{1}{\sqrt2} \quad\Longrightarrow\quad H - 2 = \frac{H}{\sqrt2} \quad\Longrightarrow\quad H\left(1 - \frac{1}{\sqrt2}\right) = 2.
Solving,
H=21−12=222−1=22(2+1)(2−1)(2+1)=4+22.H = \frac{2}{1 - \frac{1}{\sqrt2}} = \frac{2\sqrt2}{\sqrt2 - 1} = \frac{2\sqrt2(\sqrt2 + 1)}{(\sqrt2 - 1)(\sqrt2 + 1)} = 4 + 2\sqrt2.

The answer is (E) 4+22\boxed{\textbf{(E)}\ 4+2\sqrt{2}}.

Why this works

Similar solids obey the k:k2:k3k : k^2 : k^3 rule for lengths, areas and volumes. "Half the surface area" therefore means lengths shrink by 1/21/\sqrt2, and the altitude is one such length. The one number given (22) is a difference of two altitudes, so express it as H−kHH - kH.

Alternative approach

Check the choices: the small altitude must be H−2H - 2 with (H−2)/H=1/2≈0.707(H-2)/H = 1/\sqrt2 \approx 0.707. For H=4+22≈6.83H = 4 + 2\sqrt2 \approx 6.83, (H−2)/H≈4.83/6.83≈0.707(H-2)/H \approx 4.83/6.83 \approx 0.707. For H=4H = 4 the ratio is 0.50.5, for H=2+2H = 2 + \sqrt2 it is about 0.410.41; only (E) fits.

The trap

Using the scale factor 1/2 (as if lengths halved) or 1/cube-root-2 (as if volumes halved) instead of 1/sqrt 2 for areas.

Common mistakes

  • Using the scale factor 1/2 (as if lengths halved) or 1/cube-root-2 (as if volumes halved) instead of 1/sqrt 2 for areas.
  • Taking the small pyramid's altitude to be 22 rather than H−2H - 2, which gives H=22H = 2\sqrt2.

Techniques

Set up the equation/formula and compute; no special trick needed · Substitute to simplify (u = x+1/x, shifting, scaling)