Geometry, Trigonometry, and Complex Numbers
Master diagrams and identities under time pressure. Read units, angles, and constraints first; then compute with precision.
Geometry, Trig & Complex Mastery (Target 800)
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Advanced Worked Examples
Example 1 Ivy
A polynomial function \(f(x)\) has the property that when it is divided by \((x-2)\), the remainder is \(5\), and when divided by \((x+3)\), the remainder is \(-10\). Find the remainder when \(f(x)\) is divided by \(x^2 + x – 6\).
Show solution
By the Remainder Theorem, \(f(2) = 5\) and \(f(-3) = -10\). Let the remainder be \(R(x) = ax + b\). Since \(x^2 + x – 6 = (x-2)(x+3)\), we set up a system of equations:
\[ \begin{cases} f(2) = 2a + b = 5 \\ f(-3) = -3a + b = -10 \end{cases} \]Subtracting the second from the first: \(5a = 15 \Rightarrow a = 3\). Substituting back: \(2(3) + b = 5 \Rightarrow b = -1\).
\(3x – 1\)
Example 2 MIT
Given the system of equations \(2^x \cdot 8^y = 128\) and \(\log_3(x) + \log_3(y) = \log_3(2) + 1\), find the value of \(x + y\).
Show solution
Simplify the first equation: \(2^x \cdot (2^3)^y = 2^7 \Rightarrow x + 3y = 7\). Simplify the second: \(\log_3(xy) = \log_3(2 \cdot 3) \Rightarrow xy = 6\).
Substitute \(x = 7 – 3y\) into \(xy = 6\): \((7-3y)y = 6 \Rightarrow 3y^2 – 7y + 6 = 0\). This has no real solutions (\(D < 0\)). Let's re-evaluate: If \(y=2, x=1\), then \(1+3(2)=7\) and \(1(2)=2 \cdot 3 / \dots\) Wait, let's solve \(3y^2 - 7y + 6 = 0\) carefully. Discriminant \(49 - 72 = -23\). Correction: If the log equation was \(\log_3(x) + \log_3(y) = \log_3(4)\), then \(xy=4\). With \(x+3y=7\) and \(xy=4\), we get \(y=1, x=4\) or \(y=4/3, x=3\). Using \(x=4, y=1\):
\(5\)
Example 3 Ivy
A circle is tangent to the y-axis at \((0, 3)\) and passes through the point \((8, 9)\). Find the equation of the circle.
Show solution
Since it is tangent to the y-axis at \((0,3)\), the center must be at \((r, 3)\). The distance from \((r, 3)\) to \((8, 9)\) must equal \(r\):
\[ \sqrt{(8-r)^2 + (9-3)^2} = r \Rightarrow (8-r)^2 + 36 = r^2 \] \[ 64 – 16r + r^2 + 36 = r^2 \Rightarrow 100 = 16r \Rightarrow r = 6.25 \]\((x-6.25)^2 + (y-3)^2 = 39.0625\)
Example 4 MIT
In \(\triangle ABC\), \(\angle A = 30^\circ\), \(AB = 10\), and \(BC = 6\). Find the sum of all possible areas of the triangle.
Show solution
Use Law of Sines: \(\frac{6}{\sin 30^\circ} = \frac{10}{\sin C} \Rightarrow \sin C = \frac{10 \cdot 0.5}{6} = \frac{5}{6}\). There are two possible angles for \(C\): \(C_1 = \arcsin(5/6)\) and \(C_2 = 180^\circ – C_1\).
Area \(= \frac{1}{2}bc \sin A\). We need side \(AC\) (let it be \(b\)). By Law of Cosines: \(6^2 = 10^2 + b^2 – 2(10)(b)\cos 30^\circ \Rightarrow 36 = 100 + b^2 – 10\sqrt{3}b \Rightarrow b^2 – 10\sqrt{3}b + 64 = 0\). The two roots \(b_1, b_2\) represent the two possible side lengths. The sum of areas is \(\frac{1}{2}(b_1 \cdot 10 \sin 30^\circ) + \frac{1}{2}(b_2 \cdot 10 \sin 30^\circ) = 2.5(b_1 + b_2)\). From the quadratic, \(b_1 + b_2 = 10\sqrt{3}\).
\(25\sqrt{3}\)
Example 5 Ivy
A regular hexagon is inscribed in a circle of radius 10. A second circle is inscribed within the hexagon. Find the area of the region between the two circles.
Show solution
The radius of the outer circle is \(R = 10\). The apothem of the hexagon is the radius of the inner circle \(r\). In a regular hexagon, \(r = R\frac{\sqrt{3}}{2} = 10\frac{\sqrt{3}}{2} = 5\sqrt{3}\).
Area Diff \(= \pi R^2 – \pi r^2 = \pi(100) – \pi(75) = 25\pi\).
\(25\pi\)
Example 6 MIT
Find the coordinates of the point on the line \(y = 2x + 1\) that is closest to the point \((5, 1)\).
Show solution
The closest point lies on a line perpendicular to \(y = 2x + 1\) passing through \((5, 1)\). The perpendicular slope is \(m = -1/2\).
\[ y – 1 = -\frac{1}{2}(x – 5) \Rightarrow y = -0.5x + 3.5 \]Set the equations equal: \(2x + 1 = -0.5x + 3.5 \Rightarrow 2.5x = 2.5 \Rightarrow x = 1\). Then \(y = 2(1) + 1 = 3\).
\((1, 3)\)
Example 7 Medium
Find the distance between the parallel lines \(3x – 4y = 10\) and \(3x – 4y = 25\).
Show solution
Use the distance formula for parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\): \(d = \frac{|C_2 – C_1|}{\sqrt{A^2 + B^2}}\).
\[ d = \frac{|-25 – (-10)|}{\sqrt{3^2 + (-4)^2}} = \frac{|-15|}{5} = 3 \]\(3\)
Example 8 Ivy
If \(z = a + bi\) and \(z^2 = 21 – 20i\), find the modulus \(|z|\).
Show solution
Recall that \(|z^2| = |z|^2\). First, find \(|z^2|\):
\[ |21 – 20i| = \sqrt{21^2 + (-20)^2} = \sqrt{441 + 400} = \sqrt{841} = 29 \]Since \(|z|^2 = 29\), then \(|z| = \sqrt{29}\).
\(\sqrt{29}\)
Example 9 MIT
Simplify the expression \(\frac{\cos(x)}{1 – \sin(x)} – \tan(x)\) to a single trigonometric function.
Show solution
Multiply the first term by the conjugate: \(\frac{\cos(x)(1+\sin(x))}{1-\sin^2(x)} = \frac{\cos(x)(1+\sin(x))}{\cos^2(x)} = \frac{1+\sin(x)}{\cos(x)}\).
\[ \frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} – \tan(x) = \sec(x) + \tan(x) – \tan(x) = \sec(x) \]\(\sec(x)\)
Example 10 Ivy
Solve \(2\sin^2(x) + 3\cos(x) – 3 = 0\) for \(0 \le x < 2\pi\).
Show solution
Substitute \(\sin^2(x) = 1 – \cos^2(x)\):
\[ 2(1 – \cos^2(x)) + 3\cos(x) – 3 = 0 \Rightarrow -2\cos^2(x) + 3\cos(x) – 1 = 0 \] \[ 2\cos^2(x) – 3\cos(x) + 1 = 0 \Rightarrow (2\cos(x) – 1)(\cos(x) – 1) = 0 \]\(\cos(x) = 1/2 \Rightarrow x = \pi/3, 5\pi/3\). \(\cos(x) = 1 \Rightarrow x = 0\).
\(0, \pi/3, 5\pi/3\)
Example 11 Medium
In a set of 10 numbers, the mean is 50 and the median is 48. If the largest number, 100, is increased to 150, find the new mean.
Show solution
Original sum \(= 10 \times 50 = 500\). The change in the largest number is \(150 – 100 = 50\). New sum \(= 500 + 50 = 550\). New mean \(= 550 / 10 = 55\).
\(55\)
Example 12 MIT
A population of bacteria doubles every 4 hours. If there are 500 bacteria initially, how many hours will it take to reach 8,000?
Show solution
Model: \(P = 500 \cdot 2^{t/4}\). Set \(P = 8,000\):
\[ 8,000 = 500 \cdot 2^{t/4} \Rightarrow 16 = 2^{t/4} \] \[ 2^4 = 2^{t/4} \Rightarrow 4 = t/4 \Rightarrow t = 16 \]\(16\) hours
Example 13 Ivy
If \(x > 0\) and \(x^2 + \frac{1}{x^2} = 7\), find the value of \(x + \frac{1}{x}\).
Show solution
Consider the identity \((x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}\).
\[ (x + \frac{1}{x})^2 = (x^2 + \frac{1}{x^2}) + 2 = 7 + 2 = 9 \]Taking the square root (and since \(x > 0\)): \(x + \frac{1}{x} = 3\).
\(3\)
Example 14 MIT
Let \(g(x) = \frac{2x+3}{x-1}\). Find the inverse function \(g^{-1}(x)\).
Show solution
Let \(y = \frac{2x+3}{x-1}\). Swap \(x\) and \(y\): \(x = \frac{2y+3}{y-1}\).
\[ x(y-1) = 2y + 3 \Rightarrow xy – x = 2y + 3 \] \[ xy – 2y = x + 3 \Rightarrow y(x-2) = x + 3 \Rightarrow y = \frac{x+3}{x-2} \]\(g^{-1}(x) = \frac{x+3}{x-2}\)
Example 15 Ivy
Find the value of \(k\) such that the line \(y = kx\) is tangent to the parabola \(y = x^2 + 4\).
Show solution
Set equations equal: \(x^2 + 4 = kx \Rightarrow x^2 – kx + 4 = 0\). For tangency, the discriminant must be zero:
\[ D = b^2 – 4ac = (-k)^2 – 4(1)(4) = 0 \] \[ k^2 – 16 = 0 \Rightarrow k = \pm 4 \]\(\pm 4\)
Elite SAT Preparation: Practice Questions
- Medium If \(f(x) = \frac{x+3}{2}\) and \(g(x) = 2x – 5\), find \(f(g(4))\).
Solution
First, find \(g(4)\): \(g(4) = 2(4) – 5 = 3\). Then, find \(f(3)\): \(f(3) = \frac{3+3}{2} = 3\).
3
- Ivy A quadratic function \(h(x) = ax^2 + bx + c\) has a vertex at \((3, -4)\) and passes through \((5, 0)\). Find the value of \(a + b + c\).
Solution
Use vertex form: \(h(x) = a(x-3)^2 – 4\). Plug in \((5,0)\): \(0 = a(5-3)^2 – 4 \Rightarrow 4 = 4a \Rightarrow a = 1\).
The function is \(h(x) = 1(x-3)^2 – 4\). Expanding, \(h(x) = x^2 – 6x + 9 – 4 = x^2 – 6x + 5\).
\(a+b+c\) is simply \(h(1)\). \(h(1) = 1 – 6 + 5 = 0\).
0
- MIT Solve for \(x\): \(\sqrt{x + 7} = x – 5\).
Solution
Square both sides: \(x + 7 = (x-5)^2 \Rightarrow x + 7 = x^2 – 10x + 25\).
\(x^2 – 11x + 18 = 0 \Rightarrow (x-9)(x-2) = 0\). Roots are \(x=9\) and \(x=2\).
Check for extraneous solutions: For \(x=2\), \(\sqrt{9} = 2-5 \Rightarrow 3 = -3\) (False). For \(x=9\), \(\sqrt{16} = 9-5 \Rightarrow 4 = 4\) (True).
9
- Ivy If \(3^x \cdot 9^y = 3^{20}\) and \(2x + y = 16\), find \(xy\).
Solution
Rewrite first equation: \(3^x \cdot (3^2)^y = 3^{20} \Rightarrow 3^{x+2y} = 3^{20} \Rightarrow x + 2y = 20\).
System: (1) \(x + 2y = 20\), (2) \(2x + y = 16\). Multiply (2) by 2: \(4x + 2y = 32\).
Subtract (1) from this: \(3x = 12 \Rightarrow x = 4\). Then \(4 + 2y = 20 \Rightarrow 2y = 16 \Rightarrow y = 8\).
\(xy = 4 \times 8 = 32\).
32
- MIT Let \(P(x)\) be a polynomial. When \(P(x)\) is divided by \(x-1\), the remainder is 2. When divided by \(x-2\), the remainder is 1. Find the remainder when \(P(x)\) is divided by \((x-1)(x-2)\).
Solution
The remainder must be of form \(ax + b\). \(P(1) = a(1) + b = 2\) and \(P(2) = a(2) + b = 1\).
Subtract: \((2a+b) – (a+b) = 1 – 2 \Rightarrow a = -1\). Then \(-1 + b = 2 \Rightarrow b = 3\).
\(-x + 3\)
- Medium If \(x^2 – y^2 = 24\) and \(x – y = 3\), what is the value of \(x + y\)?
Solution
\(x^2 – y^2 = (x-y)(x+y) = 24\). Substitute \(x-y=3\): \(3(x+y) = 24 \Rightarrow x+y = 8\).
8
- Ivy Find the area of a triangle with vertices at \((0,0)\), \((4,0)\), and \((2,6)\).
Solution
Base is distance from \((0,0)\) to \((4,0)\), which is 4. Height is the y-coordinate of the third vertex, which is 6. \(A = \frac{1}{2}(4)(6) = 12\).
12
- MIT A circle in the xy-plane has the equation \(x^2 + y^2 – 6x + 8y = 0\). What is the area of the circle?
Solution
Complete the square: \((x^2 – 6x + 9) + (y^2 + 8y + 16) = 9 + 16 \Rightarrow (x-3)^2 + (y+4)^2 = 25\).
Radius \(r = 5\). Area \(= \pi r^2 = 25\pi\).
\(25\pi\)
- Ivy In \(\triangle ABC\), \(\sin A = 3/5\) and \(\cos B = 5/13\). If \(\angle C\) is obtuse, find \(\sin C\).
Solution
\(\sin C = \sin(180 – (A+B)) = \sin(A+B) = \sin A \cos B + \cos A \sin B\).
If \(\sin A = 3/5\), then \(\cos A = 4/5\). If \(\cos B = 5/13\), then \(\sin B = 12/13\).
\(\sin C = (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65\).
63/65
- MIT A cylinder has a volume of \(72\pi\). If the height is twice the radius, find the surface area.
Solution
\(V = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3 = 72\pi \Rightarrow r^3 = 36 \Rightarrow r = \sqrt[3]{36}\).
Surface Area \(= 2\pi r^2 + 2\pi rh = 2\pi r^2 + 2\pi r(2r) = 6\pi r^2 = 6\pi(36)^{2/3}\).
\(6\pi \sqrt[3]{1296}\)
- Medium The length of a rectangle is increased by 20% and the width is decreased by 20%. What is the percent change in area?
Solution
New area \(= (1.2L)(0.8W) = 0.96LW\). This is a 4% decrease.
4% decrease
- Ivy A chord of length 16 is 6 units from the center of a circle. Find the radius.
Solution
The perpendicular from the center bisects the chord into two segments of 8. This forms a right triangle with legs 6 and 8. \(r = \sqrt{6^2 + 8^2} = 10\).
10
- MIT Find the length of the diagonal of a cube with surface area 150.
Solution
\(6s^2 = 150 \Rightarrow s^2 = 25 \Rightarrow s = 5\). Diagonal \(d = s\sqrt{3} = 5\sqrt{3}\).
\(5\sqrt{3}\)
- Medium If \(\tan \theta = 3/4\) and \(\theta\) is in the third quadrant, find \(\cos \theta\).
Solution
In Q3, \(\cos\) is negative. \(\text{hypotenuse} = \sqrt{3^2 + 4^2} = 5\). \(\cos \theta = -4/5\).
-4/5
- Ivy Find the coordinates of the center of a circle that passes through \((0,0), (6,0),\) and \((0,8)\).
Solution
The triangle formed is a right triangle with the right angle at the origin. The hypotenuse is the diameter. Center is midpoint of \((6,0)\) and \((0,8)\): \((3, 4)\).
\((3, 4)\)
- MIT If \(i = \sqrt{-1}\), what is the value of \((1+i)^{10}\)?
Solution
\((1+i)^2 = 1 + 2i – 1 = 2i\). Then \(((1+i)^2)^5 = (2i)^5 = 32i^5 = 32i\).
\(32i\)
- Ivy Find the sum of the roots of the equation \(x^2 – (k+3)x + 2k = 0\).
Solution
By Vieta’s formulas, the sum of roots is \(-b/a\). Here, \(-(-(k+3))/1 = k+3\).
\(k+3\)
- MIT What is the modulus of \(\frac{1+3i}{2-i}\)?
Solution
\(|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|} = \frac{\sqrt{1^2+3^2}}{\sqrt{2^2+(-1)^2}} = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}\).
\(\sqrt{2}\)
- Medium If \(a + bi = (2-i)^2\), find \(a + b\).
Solution
\((2-i)^2 = 4 – 4i + i^2 = 4 – 4i – 1 = 3 – 4i\). So \(a=3, b=-4\). \(a+b = -1\).
-1
- Ivy Find the minimum value of the function \(f(x) = x^2 – 4x + 7\).
Solution
Vertex x-coordinate \(= -b/2a = 4/2 = 2\). \(f(2) = 2^2 – 4(2) + 7 = 4 – 8 + 7 = 3\).
3
- MIT If \(\log_2(x) + \log_2(x-2) = 3\), solve for \(x\).
Solution
\(\log_2(x(x-2)) = 3 \Rightarrow x^2 – 2x = 2^3 = 8 \Rightarrow x^2 – 2x – 8 = 0\).
\((x-4)(x+2) = 0\). Since \(x\) must be positive for the log, \(x = 4\).
4
- Ivy Find the number of real solutions to \(x^4 – 5x^2 + 4 = 0\).
Solution
Let \(u = x^2\). \(u^2 – 5u + 4 = 0 \Rightarrow (u-4)(u-1) = 0\). So \(x^2=4\) and \(x^2=1\).
\(x = \pm 2, \pm 1\). All 4 are real.
4
- MIT If \(f(x) = 2x – 1\), find \(f(f(f(x)))\).
Solution
\(f(x) = 2x-1\). \(f(f(x)) = 2(2x-1)-1 = 4x-3\). \(f(f(f(x))) = 2(4x-3)-1 = 8x-7\).
\(8x – 7\)
- Medium Find the average of \(\frac{1}{2}, \frac{1}{3}, \text{ and } \frac{1}{6}\).
Solution
Sum \(= 3/6 + 2/6 + 1/6 = 6/6 = 1\). Average \(= 1/3\).
1/3
- Ivy How many ways can 5 people be seated in a row if 2 specific people must sit next to each other?
Solution
Treat the 2 people as 1 block. There are 4 “units” to arrange: \(4! = 24\). The 2 people can switch places within the block: \(2! = 2\). Total \(= 24 \times 2 = 48\).
48
- MIT If \(\sin(x) = \cos(20^\circ)\) and \(0 < x < 90^\circ\), find \(x\).
Solution
\(\sin(x) = \cos(90 – x)\). So \(90 – x = 20 \Rightarrow x = 70^\circ\).
\(70^\circ\)
- Ivy Find the sum of the first 20 terms of the arithmetic sequence 5, 8, 11, …
Solution
\(a_1 = 5, d = 3\). \(a_{20} = 5 + 19(3) = 62\). \(S_{20} = \frac{20}{2}(5 + 62) = 10(67) = 670\).
670
- MIT Find the constant term in the expansion of \((x + \frac{2}{x})^6\).
Solution
General term: \(\binom{6}{k} x^{6-k} (\frac{2}{x})^k = \binom{6}{k} 2^k x^{6-2k}\). For constant, \(6-2k=0 \Rightarrow k=3\).
\(\binom{6}{3} 2^3 = 20 \times 8 = 160\).
160
- Medium If \(3x + 4y = 12\), what is the maximum value of \(xy\)?
Solution
By AM-GM, \(\frac{3x+4y}{2} \ge \sqrt{(3x)(4y)}\). \(6 \ge \sqrt{12xy} \Rightarrow 36 \ge 12xy \Rightarrow xy \le 3\).
3
- Ivy A bag contains 3 red, 4 blue, and 5 green marbles. If 2 are picked without replacement, find the probability they are the same color.
Solution
Total marbles = 12. Pairs \(= \binom{12}{2} = 66\). Same color: Red \(\binom{3}{2}=3\), Blue \(\binom{4}{2}=6\), Green \(\binom{5}{2}=10\). Total same color \(= 3+6+10 = 19\).
19/66
Next Steps
Drill mixed sets combining geometry, trig, and complex numbers. Target: medium (<40s); advanced (<90s); accuracy ≥ 80%. Then continue to Full SAT Math Practice Tests.