2002 Pan African MO Problems/Problem 5
Problem
Let be an acute angled triangle. The circle with diameter
intersects the sides
and
at points
and
respectively. The tangents drawn to the circle through
and
intersect at
.
Show that
lies on the altitude through the vertex
.
Solution
Draw lines
and
, where
and
are on
and
, respectively. Because
and
are tangents as well as
and
,
and
. Additionally, because
and
are tangents,
.
Let and
. By the Base Angle Theorem,
and
. Additionally, from the property of tangent lines,
,
,
, and
. Thus, by the Angle Addition Postulate,
and
. Thus,
and
, so
. Since the sum of the angles in a quadrilateral is 360 degrees,
. Additionally, by the Vertical Angle Theorem,
and
. Thus,
.
Now we need to prove that
is the center of a circle that passes through
. Extend line
, and draw point
not on
such that
is on the circle with
. By the Triangle Angle Sum Theorem and Base Angle Theorem,
. Additionally, note that
, and since
,
. Thus, by the Base Angle Converse,
. Furthermore,
. Therefore,
is the diameter of the circle, making
the radius of the circle. Since
is a point on the circle,
.
Thus, by the Base Angle Theorem, , so
. Since
, by the Alternating Interior Angle Converse,
. Therefore, since
,
, and
must be on the altitude of
that is through vertex
.
Solution 2 (by duck_master)
Let be the intersection of
and
. Note that
, and similarly
. Thusly,
is a cyclic quadrilateral, and
is the diameter of its circumcircle.
Next, let be the intersection of
and
; we claim that
. Note that
, so
is cyclic. Then
, so
.
Furthermore, we claim that is the midpoint of
. To show this, we use the method of phantom points: we let
be the midpoint of
. Then
, and
. Since the two values match, we have
. Similarly, we show that
. This necessarily implies
.
Finally, we show that lies on the height from
to
. Since
, we know that
is the height from
to
. But
, so
lies on
and we are done.
See Also
2002 Pan African MO (Problems) | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All Pan African MO Problems and Solutions |