The tangent of the circumcircle of triangle WXY at X is drawn, and the line through W that is parallel to the tangent intesects XY by Z. If XY = 14 and WX = 6, find YZ.
(No, the answer isn't 66/7!)
Let O be the center of the circumcircle of triangle WXY, and let T be the point where the tangent at X intersects the circumcircle.
Since the tangent at X is perpendicular to the radius OX, we have OT ⊥ XT. Also, since WX is a chord of the circumcircle, we have OT ⊥ WX. Therefore, OT is perpendicular to the line through W that is parallel to the tangent, and so the line through W that is parallel to the tangent passes through the midpoint M of OT.
Since M is the midpoint of OT, we have OM = MT. Also, since OX is a radius of the circumcircle, we have OM = OX. Therefore, MT = OX = 6.
Let P be the point where the line through W that is parallel to the tangent intersects XT. Then, since the line through W is parallel to the tangent, we have ∠XWP = ∠XTX = 90°.
Since ∠XWP = 90°, we have WP ⊥ WX. Also, since MT = 6, we have PT = MT = 6. Therefore, by the Pythagorean Theorem, we have WP = √(XT^2 - PT^2) = √(XT^2 - 36).
Since WP ⊥ WX, we have ∠WPX = 90°. Therefore, by the Pythagorean Theorem, we have WX^2 + XP^2 = WP^2. Substituting WP = √(XT^2 - 36), we get:
6^2 + XP^2 = (XT^2 - 36) XP^2 = XT^2 - 36 - 6^2 XP^2 = XT^2 - 72
Since T is on the circumcircle, we have OT = r, where r is the radius of the circumcircle. Also, since OT ⊥ XT, we have OP = PT = 6. Therefore, by the Pythagorean Theorem, we have OT^2 = OP^2 + PT^2 = 6^2 + 6^2 = 72.
Now, let Q be the point where the line through W that is parallel to the tangent intersects the line through T that is perpendicular to WX. Then, since ∠XWP = 90° and ∠XTW = 90°, we have ∠WXT = ∠XPQ. Therefore, triangles WXT and XPQ are similar, so we have:
XP/XQ = WX/WT XP/(XQ + WX) = WX/OT XP/(XQ + 6) = 6/√72 XP/(XQ + 6) = √2/2
Solving for XP, we get:
XP = (√2/2)(XQ + 6)
Substituting into the previous equation, we get:
(√2/2)(XQ + 6)^2 = XT^2 - 72
Expanding and simplifying, we get:
XQ^2 + 12XQ - 2XT^2 + 144 = 0
Substituting XT = 6√2, we get:
XQ^2 + 12XQ - 72 = 0
Using the quadratic formula, we get:
XQ = (-12 ± √(12^2 + 4(72)))/2 XQ = (-12 ± 6√5)/2
Since XQ > 0, we have:
XQ = -6 + 3√5
Finally, using the Pythagorean Theorem on YTZ, we get
YZ = 8 + 2√5
+397 The triangles XWZ and XYW (in that order) are similar.
( The angles XWZ and XYW are both equal to the angle between the tangent and the line XW,
XWZ because of the parallel lines, and XYW by the circle property, angle in the opposite segment.
Angle WXY is common to both triangles.)
So,
XZ/XW = XW/XY,
XZ = XW^2/XY = 6^2/14 = 36/14 = 18/7.
Then,
YZ = 14 -18/7 = 80/7.
The 66/7 mentioned at the bottom of the question refers to an earlier posting.
A version of this question, it had XW = 8 rather than 6, was posted July(?)/ August 2020.
