Difference between revisions of "2024 AMC 12B Problems/Problem 23"
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From symmetry, we know that <math>\overline{AV} = \overline{DV}</math>, therefore <math>\triangle{AVD}</math> is a 45-45-90 triangle. Denote <math>\overline{AV}</math> as <math>x</math> so that <math>\overline{AD} = x\sqrt{2}</math>. Doing some geometry on the isosceles trapezoid <math>ABCD</math> (we know this from the fact that it is a regular octagon) reveals that <math>\overline{AD}=1+2(\sqrt{2}/2)=1+\sqrt{2}</math> and <math>\overline{AV}=(\overline{AD})/\sqrt{2}=(\sqrt{2}+2)/2</math>. | From symmetry, we know that <math>\overline{AV} = \overline{DV}</math>, therefore <math>\triangle{AVD}</math> is a 45-45-90 triangle. Denote <math>\overline{AV}</math> as <math>x</math> so that <math>\overline{AD} = x\sqrt{2}</math>. Doing some geometry on the isosceles trapezoid <math>ABCD</math> (we know this from the fact that it is a regular octagon) reveals that <math>\overline{AD}=1+2(\sqrt{2}/2)=1+\sqrt{2}</math> and <math>\overline{AV}=(\overline{AD})/\sqrt{2}=(\sqrt{2}+2)/2</math>. | ||
− | To find the length <math>\overline{IA}</math>, we cut the octagon into 8 triangles, each | + | To find the length <math>\overline{IA}</math>, we cut the octagon into 8 triangles, each with a smallest angle of 45 degrees. Using the law of cosines on <math>\triangle{AIB}</math> we find that <math>{\overline{IA}}^2=(2+\sqrt{2})/2</math>. |
Finally, using the pythagorean theorem, we can find that <math>{\overline{IV}}^2={\overline{AV}}^2-{\overline{IA}}^2= {((\sqrt{2}+2)/2)}^2 - (2+\sqrt{2})/2 = \boxed{(1+\sqrt{2})/2}</math> which is answer choice <math>\boxed{B}</math>. | Finally, using the pythagorean theorem, we can find that <math>{\overline{IV}}^2={\overline{AV}}^2-{\overline{IA}}^2= {((\sqrt{2}+2)/2)}^2 - (2+\sqrt{2})/2 = \boxed{(1+\sqrt{2})/2}</math> which is answer choice <math>\boxed{B}</math>. | ||
+ | ~username2333 | ||
==Solution 2 (Less computation)== | ==Solution 2 (Less computation)== |
Revision as of 05:51, 14 November 2024
Problem
A right pyramid has regular octagon with side length
as its base and apex
Segments
and
are perpendicular. What is the square of the height of the pyramid?
Solution 1
To find the height of the pyramid, we need the length from the center of the octagon (denote as ) to its vertices and the length of AV.
From symmetry, we know that , therefore
is a 45-45-90 triangle. Denote
as
so that
. Doing some geometry on the isosceles trapezoid
(we know this from the fact that it is a regular octagon) reveals that
and
.
To find the length , we cut the octagon into 8 triangles, each with a smallest angle of 45 degrees. Using the law of cosines on
we find that
.
Finally, using the pythagorean theorem, we can find that which is answer choice
.
~username2333
Solution 2 (Less computation)
Let be the center of the regular octagon. Connect
, and let
be the midpoint of line segment
. It is easy to see that
and
. Hence,
Hence, the answer is
.
~tsun26