answer:1. **Define the necessary points and circles:** - Let ( O ) be the center of the circumcircle (Omega) of (triangle ATL). - Let (Gamma) be the circle (odot(AH)). - Let ( M ) be the midpoint of segments ( BC ) and ( DK ). - Let ( Y ) be a point on (Omega) such that (overline{HY}) and (overline{HA}) are tangent to (Omega). - Let ( X = overline{AY} cap overline{OH} ) (which lies on (Gamma)). - Let ( Z ) be the antipode of ( A ) with respect to (Omega). - Let ( Q = Omega cap Gamma ne A ) (which lies on (overline{HZ})). - Let ( R = overline{XX} cap overline{HH}). 2. **Prove that (overline{AO} parallel overline{BC}):** - Note that (angle ALT = angle HAT = 90^circ - angle B), so (overline{AH}) is tangent to (Omega) at ( A ). - This implies that (overline{AO} parallel overline{BC}). 3. **Prove that ( Q ) is the desired tangency point and ( Q in overline{HK}):** - Consider the cross-ratio ((A, Y; L, T)_{Omega}): [ -1 = (A, Y; L, T)_{Omega} stackrel{A}{=} (H, X; E, F)_{Gamma} stackrel{H}{=} (infty_{overline{BC}}, overline{HX} cap overline{BC}; B, C) ] - This implies that (overline{HX} cap overline{BC} equiv M). - By homothety at ( H ), points ( Z, H, K ) are collinear, so ( Q in overline{HK} ). 4. **Prove that (Omega) and (Gamma) are orthogonal:** - Note that (Omega) and (Gamma) are orthogonal, so tangents at ( A ) and ( Q ) to (Omega) concur at ( O ). - Thus, ((H, X; A, Q)_{Gamma} = -1), which implies ( R in overline{AQ} ). - Also, since ((H, X; E, F)_{Gamma} = -1), ( R in overline{EF} ). 5. **Prove that (odot(RH)) passes through ( P ) and is tangent to (overline{AH}):** - Note that (odot(RH)) passes through ( P ) and is also tangent to (overline{AH}), so (omega equiv odot(RH)). - A homothety at ( Q ) sends (Omega) to (omega). This completes the proof of the problem. (blacksquare)

answer:1. Let plane AD_1 be alpha and plane AC be beta, then AD is l. If we consider AB as m, m perp l, but m is within plane beta, hence condition 1 does not satisfy the requirement. 2. If alpha, beta, and gamma are mutually perpendicular, then we can conclude m perp beta. However, there is no condition alpha perp beta mentioned, so a counterexample could only exist here. Let plane AD_1 be alpha, plane BB_1D_1D be beta, and plane AC be gamma, then AD is m, but m forms a 45° angle with beta, hence condition 2 does not satisfy the requirement. 3. Noting that m perp alpha, unless alpha and beta are not parallel, we cannot conclude m perp beta. Let plane AD_1 be alpha, plane BB_1D_1D be beta, and plane AC be gamma, considering AB as m, but m forms a 45° angle with beta, hence condition 3 does not satisfy the requirement. 4. From n perp alpha and n perp beta, we get alpha parallel beta, and from m perp alpha, we get m perp beta; hence only condition 4 satisfies the requirement. Therefore, the answer is: boxed{4}.

answer:- Statement ①: Edge AB is perpendicular to the line containing PD. Since ABCD is a rectangle with AB and CD as opposite sides, it implies that AB is parallel to CD. We know that PA is perpendicular to the plane ABCD and hence perpendicular to any line in that plane, in particular, to CD. From this, we can conclude that AB is also perpendicular to PD because a line parallel to anther line perpendicular to a third line is also perpendicular to the third line. Thus, this statement is **correct**. boxed{text{Statement ① is correct}} - Statement ②: Plane PBC is perpendicular to plane ABCD. By definition, if plane PBC were perpendicular to plane ABCD, then line PC, which is contained in both planes, would be perpendicular to plane ABCD. We know this to be true because PA is perpendicular to plane ABCD, and it's given that PA and PC intersect at P. Therefore, every line through point P in plane PBC is also perpendicular to plane ABCD. This contradicts our earlier observation that edge PB is not perpendicular to the plane; thus, this statement must be **incorrect**. boxed{text{Statement ② is incorrect}} - Statement ③: The area of triangle PCD is greater than the area of triangle PAB. Since ABCD is a rectangle, triangles PAB and PCD share the same vertical height from vertex P to the base plane ABCD. Since E and F are midpoints of PC and PD respectively, line EF, as a midsegment in triangle PCD, is parallel to base CD and equal to half its length. This means that the base CD of triangle PCD is greater than the base AB of triangle PAB. As the triangles share the same height, the area of triangle PCD, which depends on the base length as well as the height, is greater than that of triangle PAB. Hence, this statement is **correct**. boxed{text{Statement ③ is correct}} - Statement ④: Line AE and line BF are skew lines. In three-dimensional space, skew lines are two lines that do not intersect and are not parallel. AE is in the plane containing face APC, and BF is in the plane containing face BPD. Since planes APC and BPD only intersect along PB and lines AE and BF do not intersect at any point (nor are they parallel because they lie in different planes that intersect along line PB), they are indeed skew lines. Thus, this statement is **incorrect** because a correct statement was asked for. boxed{text{Statement ④ is incorrect}}

answer:The problem asks to expand the expression using the distributive property. Here, each term within the parentheses is multiplied by 4x^3: begin{align*} (5x^2 - 3) cdot 4x^3 &= 5x^2 cdot 4x^3 - 3 cdot 4x^3 &= 20x^5 - 12x^3. end{align*} Thus, the expanded form of the expression is boxed{20x^5 - 12x^3}.