Difference between revisions of "2021 JMC 10 Problems/Problem 24"
Skyscraper (talk | contribs) (Created page with "==Problem== In cyclic convex hexagon <math>AZBXCY,</math> diagonals <math>\overline{AX}</math>, <math>\overline{BY}</math>, and <math>\overline{CZ}</math> concur at the circu...") |
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− | Using analogous reasoning and summing congruent areas, we see that <math>[\triangle{ABC}]=\ | + | Using analogous reasoning and summing congruent areas, we see that <math>[\triangle{ABC}]=\tfrac{[AZBXCY]}{2} =1</math>. Similarly, <math>[\triangle{XYZ}]=1</math>. Notice that <math>[\triangle{ABC}]+[\triangle{XYZ}]</math> counts the central hexagon two times and the smaller triangles sharing a side with that hexagon once. Therefore, adding in one copy of <math>[AZBXCY]</math> will equate to the sum of all three rectangles; |
<cmath>[AZBXCY] + [\triangle{ABC}] +[\triangle{XYZ}] = 4 = [BCYZ] +[CAZX] +[ABXY],</cmath> so we have <math>[CAZX] +[ABXY]=3,</math> as desired. | <cmath>[AZBXCY] + [\triangle{ABC}] +[\triangle{XYZ}] = 4 = [BCYZ] +[CAZX] +[ABXY],</cmath> so we have <math>[CAZX] +[ABXY]=3,</math> as desired. | ||
Latest revision as of 16:35, 1 April 2021
Problem
In cyclic convex hexagon diagonals
,
, and
concur at the circumcenter of the hexagon, and quadrilateral
has area
If the sum of the areas of
and the original hexagon is equal to
what is the sum of the areas of quadrilaterals
and
Solution
Note that is equivalent to counting the central hexagon three times, the smaller triangles sharing a side with that hexagon twice, and the outer triangles just once. If
is the orthocenter of triangle
, it is well-known that
and
are reflections of each other across the midpoint of
. Therefore,
and
are congruent.
Using analogous reasoning and summing congruent areas, we see that . Similarly,
. Notice that
counts the central hexagon two times and the smaller triangles sharing a side with that hexagon once. Therefore, adding in one copy of
will equate to the sum of all three rectangles;
so we have
as desired.