2002 Pan African MO Problems/Problem 5
Let be an acute angled triangle. The circle with diameter AB intersects the sides AC and BC at points E and F respectively. The tangents drawn to the circle through E and F intersect at P. Show that P lies on the altitude through the vertex C.
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)
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.
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