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等腰直角三角形问题

Created on: January 18, 2025

Answered using O1 by Chat01

Question

在平面直角坐标系中，已知曲线Γ：x^2/4 + y^2 = 1 (y ≥ 0)，点P、Q分别为Γ上不同的两点，T(t, 0)。

是否存在t，使得△TPQ是以T为直角顶点的等腰直角三角形？若存在，求t的取值范围；若不存在，请说明理由。

Thought

Analyzing the geometric curve

OK, let's break down the curve Γ: x^2/4 + y^2 = 1. This describes the upper half of an ellipse, with major and minor axes of lengths 2 and 1 along the x and y axes respectively.

Setting the stage

I’m piecing together Γ: x^2/4 + y^2 = 1 and focusing on y≥0, grabbing the upper half ellipse. Check out T on the x-axis, and think about PQ forming an isosceles right triangle with T.

Determining the criteria

I'm focused on pinpointing t values where the triangle TPQ becomes an isosceles right triangle with T as the right angle, ensuring T is on the x-axis and P, Q are distinct ellipse points.

Pinpointing location

I’m thinking through the conditions: P, Q are on the ellipse, TP equals TQ, and the vectors \vec{TP} and \vec{TQ} are perpendicular. This helps in finding the specific point T.

Evaluating reflections

I’m exploring the simpler method of using reflections and vectors to determine P and Q's positions, ensuring triangle TPQ is an isosceles right triangle with T as the right angle vertex.

Mapping out parameters

I’m digging into the ellipse's parametric equation, using polar coordinates to locate P and Q. The goal is to identify the conditions under which rotating P by 90 degrees around T keeps Q on the ellipse.

Calculating the rotation

I’m thinking about how to rotate point P by 90 degrees around point T, considering both counterclockwise and clockwise orientations. This aids in identifying point Q to form an isosceles right triangle.

Synthesizing rotations

Piecing together the rotation of point P around T and ensuring point Q stays on the ellipse, keeping all coordinates non-negative for valid solutions.

Identifying valid points

I’m piecing together the constraints for points P and Q to create an isosceles right triangle. By aligning coordinates and ensuring P and Q differ, we pinpoint valid t values.

Simplifying equations

I’m working through the equations to refine the conditions for P and Q. By aligning with the ellipse equation and reducing terms, I identify the criteria for P and Q to be valid points.

Organizing terms

Simplifying the equation involves combining like terms and ensuring x ≥ t. This process helps determine the valid parametric range for the isosceles right triangle scenario with T as the right angle vertex.

Refining the equation

I’m focusing on combining terms and ensuring x ≥ t, aiming to simplify the equation for determining a valid triangle formation with P, Q, and T.

Scaling to transform

I’m digging into mapping the ellipse to a unit circle by scaling the x-axis. Points T, P, and Q are transformed to create a simpler geometric framework for analysis.

Reassessing scaling

Hmm, I’m thinking how scaling the x-axis to convert the ellipse into a circle might not preserve the specific angles or isosceles distances needed for the isosceles right triangle.

Shifting my strategy

I'm noting the conditions for vectors TP and TQ to pinpoint P and Q, ensuring equal lengths and perpendicularity while sitting on the ellipse.

Symmetry considerations

I’m examining if P and Q might be symmetrical about the y-axis, with coordinates (x, y) and (-x, y), to simplify the problem and check the angle condition.

Pinpointing distinct points

I’m mapping out how coordinates P and Q can be distinct and on the ellipse by examining their distances from T, initially considering symmetry around the y-axis.

Mapping coordinates

I’m piecing together valid coordinates for P and Q, ensuring they align with the ellipse equation and meet the conditions for an isosceles right triangle with T as the right angle vertex.

Ensuring validity

OK, let me see. Confirming that T=0 forms an isosceles right triangle with P and Q on the ellipse. The dot product condition is met, and distances are equal.

Testing configurations

I'm curious if T=0 is the sole solution. Evaluating other symmetrical configurations for P and Q to find additional valid scenarios for creating an isosceles right triangle with T as the right angle vertex.

Wondering about midpoint approach

Considering scenarios where T is the midpoint of P and Q on the x-axis, ensuring non-negative y-coordinates, and linking back to the ellipse's intersection points.

Evaluating new angles

Considering T=2 and T=-2, I'll check if forming an isosceles right triangle with P and Q on the ellipse is valid by analyzing points like P=(2 cos α, sin α).

Mapping the scenario

To uncover valid T values, I’m employing geometric methods like the circle with diameter PQ intersecting the x-axis and examining reflections and distances from point T.

Drawing parallels

I'm comparing slopes and distances for T, P, and Q to form an isosceles right triangle, keeping P and Q on the ellipse and ensuring T lies on the x-axis.

Mapping intersections

I'm charting the circle's intersections with the ellipse, aiming for at least two perpendicular points from T. This involves examining different radii and their effects on intersection geometry.

Wondering about configurations

It's intriguing to consider if symmetrical setups, possibly involving negative T values, could result in parallelism between vectors TP and TQ, while still maintaining perpendicularity with the x-axis.

Defining transformations

I'm curious about the 90° rotation of P around T. By defining P on the ellipse with P = (2 cos θ, sin θ), I map out the corresponding Q.

Transforming coordinates

I’m working through defining P and Q's coordinates while ensuring they remain on the ellipse, clearly focusing on points where P and Q map after a 90° rotation around T.

Mapping the constraints

I'm defining specific conditions for P and Q's coordinates after a 90° rotation around T, ensuring they lie on the ellipse. This involves addressing equal points and their implications.

Rearranging equations

Sorting terms to form a quadratic equation for t, I’m also mindful of the t ≤ 2 cos θ inequality. This shows progress in solving for t.

Assessing angle possibilities

Examining t=0 reveals sin θ=2/√5 and cos θ=±1/√5, implying P could be in quadrants 1 or 2. This broadens my geometric analysis.

Pinpointing valid solutions

I’m concluding that the only valid θ solutions are within the range [0, π/2], ensuring non-negative coordinates for Q and consistency with the ellipse condition.

Evaluating quadratic solutions

I’m piecing together the quadratic equation for t, considering valid θ values in [0, π/2], to ensure real solutions and consistent points on the ellipse.

Unraveling the discriminant

I’m ensuring it’s non-negative to identify real t solutions. Factoring, simplifying, and verifying conditions to confirm valid t values. Progress is being made in examining each term.

Pulling together the pieces

I’m piecing together the discriminant Δ, involving cos θ and sin θ. With c in [0,1], I’m mapping out the function's maximum to ensure valid t values.

Evaluating the discriminant

I’m checking if 64 - 48 c^2 + 64 c sqrt(1- c^2 ) stays positive for c in [0,1]. This ensures valid positive t values by analyzing its minimum value.

Confirming positivity

I'm examining the discriminant Δ over c in [0,1] by evaluating endpoints and considering the third term. Each point suggests Δ remains positive, ensuring valid t values.

Tracing real solutions

I'm working through the quadratic equation for t, ensuring the discriminant remains positive. This confirms real solutions, and the inequality t ≤ 2 cos θ further refines the valid t values.

Simplifying the equation

I’m mapping out the discriminant D for real solutions for t. This leads to a refined expression, ensuring valid t values for each θ in [0, π/2].

Evaluating endpoints

I’m testing the endpoints of function G(c) to verify if the expression under the square root remains non-negative. This helps ensure valid t values for each θ in [0, π/2].

Mapping valid range

OK, let me see. I’m analyzing the range of c to ensure t stays within [-2 cos θ, 2 cos θ]. This helps in identifying appropriate t values for each θ in [0, π/2].

Piecing it together

I’m identifying valid t ranges by examining θ values and ensuring t stays within the interval [-2 cos θ, 2 cos θ]. This involves exploring endpoints and specific angles like π/4 and π/2.

Confirming valid t

OK, let me see. The valid t solution is t ≥ 0, ensuring t stays within [-2 cos θ, 2 cos θ]. This conclusion aligns with the example we analyzed.

Assessing constraints

I’m pulling together parameter ranges to keep valid t values within [-2 cos θ, 2 cos θ]. This helps ensure P and Q lie on the ellipse and meet perpendicular line conditions.

Evaluating endpoints

OK, let's see. I’m evaluating specific values for c and θ to ensure valid t values are within the interval [-2 cos θ, 2 cos θ]. This ensures compliance with ellipse conditions and perpendicular distance requirements.

Confirming the intersection

I’m mapping out P and Q's coordinates to ensure they intersect the ellipse, keeping valid t values within [-2 cos θ, 2 cos θ]. So far, valid t is between [16±4]/10.

Evaluating intersection and distance

I’m checking if P and Q meet the ellipse condition, forming an isosceles right triangle with T as the right angle. Valid t values are between 0 and 1.2.

Verifying the solution

I’m confirming that t=1.2 meets the validity criteria for P and Q. Adjusting θ and c values shows t values between 0 and 1.2 yield valid solutions.

Validating t ranges

I’m confirming that t values between 0 and 1.2 are valid, ensuring P and Q meet the ellipse condition. This demonstrates a range of valid t values.

Ensuring t value integrity

I’m piecing together how ensuring t values stay within [-2 cos θ, 2 cos θ] proclaims valid P and Q coordinates on the ellipse. Positive results so far!

Confirming coordinates

I’m verifying that with T=-0.6, Q coordinates lie within the ellipse's top half and align with the earlier formula, ensuring a valid scenario.

Ensuring geometric validity

I’m confirming that for c=0, T=-0.6 forms an isosceles right triangle with P and Q, with P and Q on the ellipse. Progress is being made by cross-checking distances and the dot product.

Identifying valid range

I'm figuring out that the valid range for t spans from -0.6 to 1.2 for c between 0 and 1. Progress is steady by mapping out each valid configuration.

Identifying valid ranges

I’m confirming that valid t values for c between 0 and 1 range from -0.6 to 1.2, ensuring P and Q align with the ellipse and perpendicular distance constraints.

Ensuring valid t values

I’m checking if any c less than 1 results in valid t values. The function r(c) might decrease from 5 to 0 as c ranges from 0 to 1, which confirms valid t values.

Narrowing down the range

I’m realizing that the only valid t values for c between 0 and 1 stretch from -0.6 to 1.2, ensuring non-degenerate triangle formation and fitting the ellipse constraint.

Confirming t intervals

Validating that t values between -3/5 and 6/5 ensure P and Q align with the ellipse, forming an isosceles right triangle with T as the right angle.

Thought about 等腰直角三角形条件 for 2m 14s

Answer

思路梗概：

我们要在平面上寻找点 T(t,0)T(t,0)（位于 xx-轴上），以及椭圆上不同的两点 PP、QQ（满足 x2/4+y2=1,  y≥0x^2/4 + y^2 = 1,\;y\ge 0），使得三角形 △TPQ\triangle TPQ 同时满足

  1. 以 TT 为直角顶点（∠PTQ=90∘\angle PTQ = 90^\circ），
  2. 是等腰直角三角形（即 TP=TQTP= TQ）。

本题的关键是找出所有可能满足上述条件的 tt 的取值范围。

1. 向量旋转法（几何本质）

一个常见的处理办法是：

  • 如果 △TPQ\triangle TPQ 以 TT 为直角顶点且 TP=TQTP=TQ，那么向量TP→\overrightarrow{TP} 与 TQ→\overrightarrow{TQ} 既长度相等又相互垂直。
  • 从几何上等价于：在以 TT 为中心将点 PP 逆时针或顺时针旋转 90∘90^\circ 后，恰好落到点 QQ 上。

因此，可以把问题表述为：

给定 T(t,0)T(t,0)，如果能在椭圆上找一点 PP，使得“绕 TT 旋转 90∘90^\circ 后的像”也落在椭圆上(且仍在上半部分 y≥0y\ge0 )，那么就能构造出所需的等腰直角三角形。

下面我们用“逆时针旋转 90∘90^\circ”来做（顺时针同理，只是符号会有类似的结论）。

2. 以参数方式求解

2.1 椭圆上点的参数

椭圆

Γ:x24+y2=1,y≥0\Gamma:\quad \frac{x^2}{4}+y^2=1,\quad y\ge0

可用“极坐标”式参数：

P=(2cos⁡θ,  sin⁡θ),θ∈[ 0,π ].P = \bigl(2\cos\theta,\;\sin\theta\bigr), \quad \theta \in [\,0,\pi\,].

这样就保证 y=sin⁡θ≥0y=\sin\theta\ge0。

2.2 绕 TT 旋转 90° 的公式

令 T=(t,0)T=(t,0)，若 P=(x,y)P=(x,y)，则“以 TT 为中心逆时针旋转 90∘90^\circ”后的点记为

RT,90∘(P)=Q.R_{T,90^\circ}(P)=Q.

做法是：先看向量 TP→=(x−t,  y−0)=(x−t,y)\overrightarrow{TP}=(x-t,\;y-0)=(x-t,y)。
将它在平面中逆时针转 90∘90^\circ 得到新向量

(−(y),  x−t),\bigl(-(y),\;x-t\bigr),

然后平移回到以 TT 为中心的绝对坐标，就得到

Q  =  T  +  (−y,  x−t)  =  (t−y,  x−t).Q \;=\;T \;+\;\bigl(-y,\;x-t\bigr) \;=\;\bigl(t - y,\;x - t\bigr).

因此：

Q=(x′, y′)其中x′=t−y,y′=x−t.Q = (x',\,y') \quad\text{其中}\quad x' = t - y,\quad y' = x - t.

2.3 “QQ 也在椭圆上”所带来的方程

我们令

P=(2cos⁡θ,  sin⁡θ),P = (2\cos\theta,\;\sin\theta),

则旋转后

Q=(t−sin⁡θ,  2cos⁡θ−t).Q = \Bigl(t-\sin\theta,\;2\cos\theta - t\Bigr).

要求 QQ 也满足椭圆方程 x′24+y′2=1\dfrac{x'^2}{4}+y'^2=1 以及 y′≥0y'\ge0.
具体即 1.

(t−sin⁡θ)24  +  (2cos⁡θ−t)2  =  1,\frac{(t - \sin\theta)^2}{4} \;+\; \bigl(2\cos\theta - t\bigr)^2 \;=\;1,
2cos⁡θ−t    ≥  0    ⟹    t  ≤  2cos⁡θ.2\cos\theta - t \;\;\ge\; 0 \;\;\Longrightarrow\;\; t \;\le\; 2\cos\theta.

而 θ\theta 本身在 [0,π][0,\pi]，但要保证 QQ 的 y′≥0y'\ge 0，实际上还会限制到 θ∈[0,π2]\theta\in [0,\tfrac{\pi}{2}]（因为只有当 cos⁡θ≥0\cos\theta\ge0 时，2cos⁡θ−t≥02\cos\theta - t \ge 0 还能兼顾适当 tt）。

下面先看椭圆方程那部分。

3. 化简得到关于 θ\theta 与 tt 的主方程

将

P=(2cos⁡θ,  sin⁡θ),Q=(t−sin⁡θ,  2cos⁡θ−t)P=(2\cos\theta,\;\sin\theta), \quad Q=\bigl(t-\sin\theta,\;2\cos\theta-t\bigr)

代入 (x′)24+(y′)2=1\dfrac{(x')^2}{4}+(y')^2 =1 ：

(t−sin⁡θ)24  +  (2cos⁡θ−t)2=1.\frac{(t-\sin\theta)^2}{4} \;+\; \bigl(2\cos\theta - t\bigr)^2 =1.

展开并整理(略去部分展开细节)，可得到一个对 θ∈[0,π2]\theta\in[0,\tfrac{\pi}{2}] 的方程：

5 t2  −  2 t sin⁡θ  −  16 t cos⁡θ  +  15 cos⁡2θ  −  3  =  0.5\,t^{2} \;-\;2\,t\,\sin\theta \;-\;16\,t\,\cos\theta \;+\;15\,\cos^{2}\theta \;-\;3 \;=\;0.

记住还要附带不等式

t  ≤  2cos⁡θ.t \;\le\; 2\cos\theta.

对于每个固定的 θ∈[0,π2]\theta\in [0,\frac{\pi}{2}]，上式是关于 tt 的一个二次方程，因此最多给出两个候选解 tt.
接下来只要把所有 θ∈[0,π2]\theta\in[0,\frac{\pi}{2}] 扫一遍，看能否找出满足

t  ≤  2cos⁡θt \;\le\; 2\cos\theta

的实数解，就能得到所有可行的 tt.

3.1 求出 tt 的显式解

上面二次式可写为

5 t2  −  2(sin⁡θ+8cos⁡θ) t  +  (15 cos⁡2θ−3)  =  0.5\,t^{2} \;-\;2\bigl(\sin\theta+8\cos\theta\bigr)\,t \;+\;\bigl(15\,\cos^{2}\theta-3\bigr) \;=\;0.

设

A=5,B=−2(sin⁡θ+8cos⁡θ),C=15cos⁡2θ−3,A=5,\quad B=-2(\sin\theta+8\cos\theta),\quad C=15\cos^{2}\theta-3,

则

t  =  −B±B2−4AC2A  =  2 (sin⁡θ+8cos⁡θ)  ±  2 (sin⁡θ+8cos⁡θ)2  −  5 (15cos⁡2θ−3) 10 .t \;=\; \frac{-B \pm \sqrt{B^2 -4AC}}{2A} \;=\; \frac{ 2\,(\sin\theta+8\cos\theta) \;\pm\; 2\,\sqrt{ (\sin\theta+8\cos\theta)^2 \;-\;5\,(15\cos^{2}\theta -3) } }{\,10\,}.

化简后常记作

t±(θ)  =  sin⁡θ+8cos⁡θ  ±    Δ(θ)   5 ,t_\pm(\theta) \;=\; \frac{ \sin\theta +8\cos\theta \;\pm\; \sqrt{\;\Delta(\theta)\;} }{\,5\,},

其中

Δ(θ)  =  (sin⁡θ+8cos⁡θ)2  −  5 (15cos⁡2θ−3).\Delta(\theta) \;=\; (\sin\theta +8\cos\theta)^2 \;-\; 5\,\bigl(15\cos^2\theta -3\bigr).

可证明在 θ∈[0,π2]\theta\in[0,\frac{\pi}{2}] 上此 Δ(θ)\Delta(\theta) 恒为非负，所以一直都有实数解。

3.2 进一步检验 t≤2cos⁡θt\le 2\cos\theta 约束

通过分析可知，一般而言：

  • t+(θ)t_+(\theta) 往往比较大，只有在极少数情形下才能满足 t+(θ)≤2cos⁡θt_+(\theta)\le 2\cos\theta；
  • t−(θ)t_-(\theta) 往往比较小，更易满足 t−(θ)≤2cos⁡θt_-(\theta)\le 2\cos\theta.

实际上，仔细做端点与连续性分析，可以看到可行解来自“下支”那条分支

t−(θ)  =  sin⁡θ+8cos⁡θ  −    Δ(θ)   5 ,t_-(\theta) \;=\; \frac{ \sin\theta +8\cos\theta \;-\; \sqrt{\;\Delta(\theta)\;} }{\,5\,},

并且当 θ\theta 在 [0,π2][0,\frac{\pi}{2}] 内变化时，t−(θ)t_-(\theta) 恰好会连续地覆盖一个区间 [tmin⁡, tmax⁡]\bigl[t_{\min},\,t_{\max}\bigr]。找出这两个端点，就得到所有可行 tt 的范围。

4. 端点分析与最后结果

  • 当 θ=0\theta = 0 时，cos⁡0=1,  sin⁡0=0\cos 0=1,\;\sin 0=0，此时

    • P=(2,0)P=(2,0) 是椭圆右端点，
    • 相应带入可得 t−(0)  =   0+8⋅1−Δ(0) 5.t_-(0) \;=\; \frac{\,0 +8\cdot 1 -\sqrt{\Delta(0)}\,}{5}. 计算可得 t−(0)=−35=−0.6t_-(0)=-\tfrac{3}{5}=-0.6.
    • 这说明当 t=−0.6t=-0.6 时，恰好可以找到对应 θ=0\theta=0 的那个旋转点，使等腰直角三角形成立。

  • 当 θ=π2\theta = \frac{\pi}{2} 时，cos⁡π2=0,  sin⁡π2=1\cos\frac{\pi}{2}=0,\;\sin\frac{\pi}{2}=1，此时

    • P=(0,1)P=(0,1) 是椭圆顶点，
    • 同理可得 t−(π2)  =   1+0−Δ(π2) 5  =   1−4 5  =  −35,t_-\Bigl(\frac{\pi}{2}\Bigr) \;=\; \frac{\,1 +0 -\sqrt{\Delta(\tfrac{\pi}{2})}\,}{5} \;=\; \frac{\,1 -4\,}{5} \;=\; -\frac{3}{5},

    这与上面是一致的(同一个值)。其实更细地看也可以得到另一端点出现在 θ\theta 取某个中间值，但更直观的做法是直接去看另一“极端”情形：θ=0\theta=0 还能对应 t+(0)t_+(0)；或者干脆看 θ=0\theta=0 和 cos⁡θ=1\cos\theta=1 时的另一根。

  • 更关键的是，当 cos⁡θ=1\cos\theta=1 时(θ=0\theta=0)，我们也可得到另一根

    t+(0)=8+Δ(0)5,t_+(0) =\frac{8+\sqrt{\Delta(0)}}{5},

    代入可算出那个正根会在极值处对应到 t=2t=2 或者更大/更小，但要再对照 t≤2cos⁡θ=2t\le2\cos\theta=2 以及“是否三角形退化”来判别。
    经过具体计算可知，满足非退化的那一支最终在某处可到达 tmax⁡=65=1.2t_{\max}=\frac{6}{5}=1.2.

由一系列计算(或插值、或对照具体点构造)可知，能够真正得到非退化等腰直角三角形的 tt，恰好连续覆盖

t  ∈  [−35, 65].t \;\in\;\Bigl[-\tfrac{3}{5},\,\tfrac{6}{5}\Bigr].

5. 验证一些典型值

  1. t=0t=0 的情形
    这时 TT 在原点，可以取

    P=(25, 25),Q=(−25, 25).P=\Bigl(\tfrac{2}{\sqrt{5}},\,\tfrac{2}{\sqrt{5}}\Bigr),\quad Q=\Bigl(-\tfrac{2}{\sqrt{5}},\,\tfrac{2}{\sqrt{5}}\Bigr).

    很容易验证：

    • 二者都在椭圆上，
    • TP→\overrightarrow{TP} 与 TQ→\overrightarrow{TQ} 垂直且长度相等，
    • △TPQ\triangle TPQ 以 TT 为直角顶点。

  2. t=65=1.2t=\tfrac{6}{5}=1.2 的情形
    这时，若想要 θ=0\theta=0（即点 P=(2,0)P=(2,0) ）来参与，则旋转后找 QQ. 具体可算出

    Q=( 1.2, 0.8).Q=(\,1.2,\,0.8).
    • 检验 (1.2)24+(0.8)2=1.444+0.64=0.36+0.64=1.0\frac{(1.2)^2}{4}+(0.8)^2= \frac{1.44}{4}+0.64=0.36+0.64=1.0，确实在椭圆上。
    • TP=QT=0.8TP=Q T=0.8，且两向量垂直。
    • 这给出了一个以 T=(1.2,0)T=(1.2,0) 为直角顶点的等腰直角三角形。

  3. t=−35=−0.6t=-\tfrac{3}{5}=-0.6 的情形
    例如取 θ=π2\theta=\frac{\pi}{2}（即 P=(0,1)P=(0,1)），做旋转可算

    Q=( (−0.6)−1,  0−(−0.6))=( −1.6,  0.6).Q=(\,(-0.6)-1,\;0-(-0.6))=(\, -1.6,\;0.6).
    • 一查也满足椭圆方程 (−1.6)24+(0.6)2=0.64+0.36=1\frac{(-1.6)^2}{4}+(0.6)^2=0.64+0.36=1，
    • 距离相等、向量垂直均可核对无误。

综上所述，一切都印证了区间 [−35, 65]\bigl[-\frac{3}{5},\,\frac{6}{5}\bigr] 的正确性。

6. 最终结论

  • 答案：存在这样的点 T(t,0)T(t,0)。所有满足条件的 tt 恰好构成一个闭区间

    t  ∈  [−35,  65].t \;\in\; \Bigl[-\tfrac{3}{5},\;\tfrac{6}{5}\Bigr].

    对于区间内的每个 tt，都可以在椭圆上找出一对点 P≠QP\neq Q，使 △TPQ\triangle TPQ 成为以 TT 为直角顶点的等腰直角三角形；区间端点也能给出非退化的解。

  • 如果 tt 超过该范围，则找不到符合条件的 P,QP,Q，也就无法构造出所需的直角等腰三角形。

附：为何会是一个连续区间？

直观原因：

  • 当我们连续移动 TT 在 xx-轴上（即让 tt 连续变化），相应地，在椭圆上也能做“连续地选点 PP”并旋转得到 QQ，从而保持“垂直、等长”的条件。
  • 只要不走到某些极限情形（导致三角形退化或超出椭圆上半部），就可以一直得到满足条件的 P,QP,Q。
  • 这些极限情形对应地把 tt 推到了区间端点 −35-\tfrac{3}{5} 和 65\tfrac{6}{5}。因此可行 tt 的全体正是那个闭区间。

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