AMC 10 · 2014 · #24
Grade 11 geometry-2dPick an answer.
Five unknown diagonals and no obvious right angles is too much to hold at once, so the first job is to shrink the unknown count. The side lengths repeat — 3,10,3,10,14 — and on a circle repeated chords mean repeated arcs, which forces three of the five diagonals to be literally the same segment length. That drops the problem to three unknowns, and then two applications of Ptolemy's theorem (the exact metric identity every cyclic quadrilateral obeys) give two equations in two of them, which collapse to one cubic. The cubic has a single positive root, so the value is forced. One trap is worth naming in advance: every equation used here is a consequence of the pentagon existing. Solving them proves only that IF the pentagon exists THEN the diagonals are these numbers — it does not produce the pentagon. So the figure gets built explicitly at the end, from the circle outward, and the two side lengths BC=10 and DE=10 are computed rather than assumed. Ptolemy is also chosen over trigonometry deliberately: it needs no picture-reading, so no fact is smuggled in from a drawing.
Repeated sides force repeated arcs
Repeated sides force repeated arcs.
Same chord, same circle, same bite of the circle — unless the two bites together swallow the whole circle, which there is no room for.
The same chord in the same circle always cuts off the same bite of the circle.
▸ Why?
Every point of the circle sits one radius from the centre, so equal chords sit at equal central angles.
▸ Why?
An arc is the share of the whole circle its central angle takes, so equal angles mean equal arcs.
Three diagonals are one length
Three diagonals share one length.
A diagonal only cares how much of the circle it jumps over, and these three jump over the same 3-arc plus 10-arc.
10.G-C.A.2Introduce A VariableTwo Ptolemy equations
Two applications of the identity give two equations.
On a circle, the two diagonals of any four points are locked to the four sides by one exact multiplication rule.
10.G-SRT.B.5Convert To AlgebraEliminate a, land on a cubic
Eliminating leaves one cubic.
Two equations in two unknowns collapse to one equation in one unknown, and a cubic with one positive root leaves no choice.
11.A-APR.B.3Convert To AlgebraBuild the pentagon before trusting the number
The pentagon really can be drawn.
Solving equations proves what the answer must be; drawing the figure proves there is an answer at all.
11.G-SRT.D.11Draw A DiagramRead off BE
One diagonal follows immediately.
Once the hard length is known, the rest of the system just hands over its answers.
8.EE.C.7Introduce A VariableOne more Ptolemy for AD
One more use of the identity gives the last.
Dropping a vertex makes a new quadrilateral whose diagonals are precisely the two lengths still in play.
10.G-SRT.B.5Convert To AlgebraAdd the five diagonals
Adding gives a fraction over six.
Three equal diagonals plus two fractions is just one common-denominator addition.
5.NF.A.1Identify SubproblemsReduce and choose
It is already reduced, giving 391.
A fraction only counts as m/n once you have checked nothing cancels.
6.NS.B.4Eliminate PossibilitiesEqual sides on a circle cut equal arcs, so three of the five diagonals are secretly the same length — name that one length and Ptolemy's rule squeezes the whole pentagon into a single cubic equation.
- Repeated sides force repeated arcs
- Three diagonals are one length
- Two Ptolemy equations
- Eliminate a, land on a cubic
- Build the pentagon before trusting the number
- Read off BE
- One more Ptolemy for AD
- Add the five diagonals
- Reduce and choose