Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 12"
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==Solution 3 (Mass points and Ptolemy)== | ==Solution 3 (Mass points and Ptolemy)== | ||
− | Let <math>O</math> be the center of square <math>BDEF</math>. Applying moment of inertia to the system of mass points <math>\Sigma = {1B,1D,1E,1F}</math> (which has center of mass <math>O</math>) gives <cmath>CB^2 + CD^2 + CE^2 + CF^2 = OB^2 + OD^2 + OE^2 + OF^2 + 4OC^2.</cmath> Since <math>\triangle CBD</math> is a right triangle, we may further cancel out some terms via Pythag to get <cmath>CE^2 + CF^2 = OE^2 + OF^2 + | + | Let <math>O</math> be the center of square <math>BDEF</math>. Applying moment of inertia to the system of mass points <math>\Sigma = {1B,1D,1E,1F}</math> (which has center of mass <math>O</math>) gives <cmath>CB^2 + CD^2 + CE^2 + CF^2 = OB^2 + OD^2 + OE^2 + OF^2 + 4OC^2.</cmath> Since <math>\triangle CBD</math> is a right triangle, we may further cancel out some terms via Pythag to get <cmath>CE^2 + CF^2 = OE^2 + OF^2 + 4OC^2 = 65 + 4OC^2.</cmath> To compute <math>OC</math>, apply Ptolemy to cyclic quadrilateral <math>DOCB</math> (using the fact that <math>\triangle BOD</math> is 45-45-90) to get <math>OC = \tfrac{3}{\sqrt 2}</math>. Thus <cmath>CE^2 + CF^2 = 65 + 4\cdot\frac 92 = 65 + 18 = \boxed{83}.</cmath> ~djmathman |
==See also== | ==See also== |
Latest revision as of 20:45, 22 December 2021
Contents
[hide]Problem
Rectangle is drawn such that
and
.
is a square that contains vertex
in its interior. Find
.
Solution 1 (Clever Construction)
We draw a line from to point
on
such that
. We then draw a line from
to point
on
such that
. Finally, we extend
to point
on
such that
.
Next, if we mark as
, we know that
, and
. We repeat this, finding
, so by AAS congruence,
. This means
, and
, so
. We see
, while
. Thus,
~Bradygho
Solution 2 (Trig)
Let . We have
, and
. Now, Law Of cosines on
and
gets
and
, so
~ Geometry285
Solution 3 (Mass points and Ptolemy)
Let be the center of square
. Applying moment of inertia to the system of mass points
(which has center of mass
) gives
Since
is a right triangle, we may further cancel out some terms via Pythag to get
To compute
, apply Ptolemy to cyclic quadrilateral
(using the fact that
is 45-45-90) to get
. Thus
~djmathman
See also
- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.