1963 AHSME Problems/Problem 39
Contents
[hide]Problem 39
In lines
and
are drawn so that
and
. Let
where
is the intersection point of
and
. Then
equals:
Solution
Draw line , and let
,
, and
, so
and
. Because
and
share an altitude,
Because
and
share an altitude,
Thus,
, and since
,
, which is answer choice
.
Solution 2 (Mass Geometry)
Let the mass of point ,
,
,
, and
be
,
,
,
, and
respectively.
By mass geometry theorems, we have
Focusing on the line segment
, using mass geometry theorems, we have
and
which leads to
.
For line segment
, similarly, we got
Substituting
and
back to the equation we obtained at the beginning, we got:
which gives us the answer choice
. -nullptr07
Solution 3 (Menelaus’s Theorem)
By using Menelaus’s Theorem on triangle BCE and points A, P, D, we can substitute in the known values to find that the answer is 5.
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 38 |
Followed by Problem 40 | |
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