Triangles — PYPs

The Setup: We need to calculate the total number of triangles that can be formed with integer sides and a fixed perimeter of $15$.

Step 1: Define the basic parameters and constraints.
Let the integer sides of the triangle be $a$, $b$, and $c$, strictly ordered such that $a \le b \le c$.
The perimeter condition dictates:
$$a + b + c = 15$$

Step 2: Apply the triangle inequality theorem to bound the longest side $c$.
The sum of the two shorter sides must strictly exceed the longest side:
$$a + b > c$$
Substitute $a + b = 15 - c$ into the inequality:
$$15 - c > c \Rightarrow 2c < 15 \Rightarrow c \le 7.5$$
Since $c$ is an integer, the maximum valid dimension for $c$ is $7$.
Furthermore, $c$ is the maximum side, so it must be at least the average length of the perimeter:
$$c \ge \frac{15}{3} = 5$$
Thus, $c \in \{5, 6, 7\}$.

Step 3: Systematically evaluate integer pairs $(a, b)$ for each possible value of $c$, maintaining $a \le b \le c$.
Case 1: $c = 5$
Requires $a + b = 10$. The only integer pair satisfying $a \le b \le 5$ is $(5, 5)$. (Yields $1$ triangle)
Case 2: $c = 6$
Requires $a + b = 9$. The pairs satisfying $a \le b \le 6$ are $(3, 6)$ and $(4, 5)$. (Yields $2$ triangles)
Case 3: $c = 7$
Requires $a + b = 8$. The pairs satisfying $a \le b \le 7$ are $(1, 7)$, $(2, 6)$, $(3, 5)$, and $(4, 4)$. (Yields $4$ triangles)

Step 4: Aggregate the valid triangle formations.
$$\text{Total Triangles} = 1 + 2 + 4 = 7$$

Final Answer: 7

The Setup: A right-angled triangle has vertex $B$ at the right angle, with legs $AB = BC = 18$. We need to find the area of the largest inscribed rectangle sharing vertex $B$.

Step 1: Establish the coordinate system.
Place the right-angled vertex $B$ at the origin $(0,0)$.
Given leg lengths $AB = BC = 18$, set vertex $A$ at $(0,18)$ and vertex $C$ at $(18,0)$.
The hypotenuse $AC$ forms a line passing through $(0,18)$ and $(18,0)$. Its linear equation is:
$$x + y = 18 \Rightarrow y = 18 - x$$

Step 2: Define the area function of the inscribed rectangle.
A rectangle sharing vertex $B$ at $(0,0)$ and bounded by the triangle will have its opposing vertex $(x, y)$ strictly on the hypotenuse $AC$.
The Area ($A$) is the product of its length and width:
$$A_{\text{rect}} = x \times y = x(18 - x) = 18x - x^2$$

Step 3: Maximize the quadratic area function.
The function is a downward-opening parabola. We find its maximum via differentiation (or the vertex formula $x = \frac{-b}{2a}$):
$$\frac{d A_{\text{rect}}}{dx} = 18 - 2x = 0 \Rightarrow x = 9$$

Step 4: Calculate the maximum area footprint.
Substitute $x = 9$ into the bounding equation to find $y$:
$$y = 18 - 9 = 9$$
$$\text{Max Area} = 9 \times 9 = 81$$

Final Answer: 81

The Setup: This geometry problem is a triple filter. The sides must pass the triangle inequality vibe check, then survive the acute-angle condition, then fit under the perimeter cap. Whichever filter is strictest wins.
Step 1: Set up the side lengths. Let the three consecutive integer sides be $x-1$, $x$ and $x+1$, with $x-1\geq 1$.
For any triangle the two smaller sides must beat the largest: $(x-1)+x>x+1\implies x>2$.
Step 2: Enforce the acute angle rule. A triangle is acute when *every* angle is acute - but we only have to test one. The largest angle always sits opposite the longest side, so if that angle is acute the two smaller ones are smaller still and acute automatically. The longest side here is $x+1$, so the single condition to check is $a^2+b^2>c^2$:

$$(x-1)^2+x^2>(x+1)^2$$

$$x^2-2x+1+x^2>x^2+2x+1$$

$$x^2>4x$$

Since $x$ is positive we may divide by $x$, giving $x>4$. As $x$ is an integer, the lowest valid value is $x=5$ (the smallest qualifying triangle being $4,5,6$). Note this condition is strictly tighter than Step 1's $x>2$, so the triangle inequality is absorbed and adds nothing further.
Step 3: Apply the perimeter cap. The perimeter is at most 100:

$$(x-1)+x+(x+1)=3x\leq 100 \implies x\leq 33.33\ldots$$

Since $x$ is an integer, the maximum is $x=33$, giving sides $32,33,34$ and a perimeter of $99$.
Step 4: Count the surviving squad. The valid range for $x$ runs from 5 to 33 inclusive:

$$33-5+1=29$$

Final Answer: 29

The Setup: We have an isosceles triangle stepping into the arena, and we need to link its side lengths, angles, and circumradius. The ultimate cheat code for any circle circumscribing a triangle is the Extended Law of Sines. It is basically the universal API bridging triangle sides, their opposite angles, and the circumcircle's radius.

Step 1: Decode the triangle's geometry.
We are given that $AB=AC=x$, making triangle $ABC$ isosceles. In geometry lore, the angle opposite to side $AC$ is $\angle ABC$, which we are told is $\theta$.

Step 2: Deploy the Extended Law of Sines.
The law dictates that the ratio of any side length to the sine of its opposite angle is exactly equal to the circumcircle's diameter (twice the circumradius, or $2R$).
The general framework is:
$$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} = 2R$$

Step 3: Plug in the stats.
Let's target side $AC$ (which has a length of $x$) and its corresponding opposite angle, $\angle ABC$ (which is $\theta$). The circumradius $R$ is given to us as $y$. Substitute these into the formula:
$$\frac{x}{\sin(\theta)} = 2y$$

Step 4: Isolate the target ratio.
The prompt is asking for the exact value of $\frac{x}{y}$. We just need to execute a quick algebraic shuffle—divide both sides by $y$ and multiply both sides by $\sin(\theta)$ to get the variables aligned perfectly.
$$\frac{x}{y} = 2\sin(\theta)$$
The math literally solves itself.

Final Answer: $2\sin(\theta)$

The Setup: A triangle is composed of side lengths $x, 21$ and $40$, where $x$ is stated to be the shortest side. We must determine a viable integer value for $x$ bounded by the triangle inequality.

Step 1: Apply the triangle inequality theorem.
For three lengths to successfully form a closed triangle, the sum of the two shorter sides must strictly exceed the longest side.

$$x + 21 > 40 \implies x > 19$$

Step 2: Apply the specific length constraint.
The problem states that $x$ is the shortest side. Therefore, $x$ must be less than or equal to $21$.
Combining both strict inequalities:

$$19 < x \le 21$$

Step 3: Evaluate the given options.
The mathematically possible integer values for $x$ are $20$ and $21$. Reviewing the multiple-choice options provided ($18, 20, 19, 16$), the only valid candidate listed is $20$.

Final Answer: 20

The Setup: In triangle $ABC$, the side $AB$ is $c$, the median $BD$ has length $k$, and the angle $\angle BDA = \theta$ where $\theta < 90^\circ$. We need to find the area of the entire triangle $ABC$.

Step 1: Establish the area formula using the median.
A median divides a triangle into two smaller triangles of identical area.
$$\text{Area}(ABC) = 2 \times \text{Area}(ABD)$$
Using the base $AD$ and the altitude from $B$ to $AD$ (which is $k\sin\theta$):
$$\text{Area}(ABD) = \frac{1}{2} \times AD \times (k\sin\theta)$$
$$\text{Area}(ABC) = 2 \left( \frac{1}{2} \times AD \times k\sin\theta \right) = AD \cdot k\sin\theta$$

Step 2: Use the Cosine Rule to express $AD$.
In $\triangle ABD$, apply the Law of Cosines:
$$c^2 = AD^2 + k^2 - 2(AD)(k)\cos\theta$$
Rearrange into a quadratic equation in terms of $AD$:
$$AD^2 - (2k\cos\theta)AD + (k^2 - c^2) = 0$$
Solve for $AD$ via the quadratic formula (taking the positive valid geometric root):
$$AD = \frac{2k\cos\theta + \sqrt{4k^2\cos^2\theta - 4(k^2 - c^2)}}{2}$$
$$AD = k\cos\theta + \sqrt{c^2 - k^2(1 - \cos^2\theta)} = k\cos\theta + \sqrt{c^2 - k^2\sin^2\theta}$$

Step 3: Substitute $AD$ back into the Area equation.
$$\text{Area}(ABC) = \left( k\cos\theta + \sqrt{c^2 - k^2\sin^2\theta} \right) k\sin\theta$$
$$\text{Area}(ABC) = k^2\sin\theta\cos\theta + k\sin\theta\sqrt{c^2 - k^2\sin^2\theta}$$
Using the double-angle identity $2\sin\theta\cos\theta = \sin(2\theta)$:
$$\text{Area}(ABC) = \frac{k^2\sin 2\theta}{2} + k\sin\theta\sqrt{c^2 - k^2\sin^2\theta}$$

Final Answer: $\dfrac{k^{2}\sin 2\theta}{2}+k\sin\theta\sqrt{c^{2}-k^{2}\sin^{2}\theta}$

The Setup: In a triangle $ABC$ with $AB=AC$, $D$ is a point on $BC$ creating $\angle BAD=30^\circ$, and $E$ is a point on $AC$ creating $AD=AE$. We need to determine the angle $\angle CDE$.

Step 1: Assign variables to the base angles of the primary triangle.
Let $\angle B = \angle C = \alpha$ (since $AB = AC$).
The vertex angle of the large triangle is $\angle BAC = 180^\circ - 2\alpha$.
The remaining upper angle inside is $\angle DAC = \angle BAC - \angle BAD = (180^\circ - 2\alpha) - 30^\circ = 150^\circ - 2\alpha$.

Step 2: Determine the base angles of the internal isosceles triangle.
In $\triangle ADE$, since $AD = AE$, the base angles are equal:
$$\angle ADE = \angle AED = \frac{180^\circ - \angle DAC}{2}$$
$$\angle ADE = \frac{180^\circ - (150^\circ - 2\alpha)}{2} = \frac{30^\circ + 2\alpha}{2} = 15^\circ + \alpha$$

Step 3: Use the exterior angle theorem to map $\angle CDE$.
Look at $\triangle ABD$. The exterior angle at $D$ is $\angle ADC$, equaling the sum of the remote interior angles:
$$\angle ADC = \angle B + \angle BAD = \alpha + 30^\circ$$
Geometrically, angle $\angle ADC$ is comprised of two adjacent components:
$$\angle ADC = \angle ADE + \angle CDE$$
Substitute the known expressions:
$$\alpha + 30^\circ = (15^\circ + \alpha) + \angle CDE$$
$$\angle CDE = 30^\circ - 15^\circ = 15^\circ$$

Final Answer: $15^{\circ}$

The Setup: This is a locus problem disguised as a dog-walking simulator. The dog maintains a strict $1$-meter social distance from the triangle at all times. When you trace this path, the dog's route is built from two types of geometry: straight paths parallel to the triangle's sides, and circular arcs pivoting around the corners. It is basically Mario Kart drifting mechanics.

Step 1: Analyze the straightaways.
For every side of the triangle, the dog runs a parallel straight line exactly $1$ meter away. The lengths of these straight paths match the triangle's base stats perfectly.
$$\text{Straight distance} = 4 + 6 + 9 = 19$$

Step 2: Calculate the corner drifts.
When the dog reaches a corner, it has to swing around the vertex to stay exactly $1$ meter away from the point. This movement creates a circular arc. The angle of this arc is exactly equal to the exterior angle of the triangle at that specific vertex.

Step 3: Piece the arcs together.
In Euclidean geometry lore, the sum of the exterior angles of *any* convex polygon is an automatic $360^\circ$ (or $2\pi$ radians). This means if you take those three corner arcs and stitch them together, they form one flawless, complete circle with a radius of $r = 1$ meter. Let's run the circumference formula:
$$\text{Arc distance} = 2\pi r = 2\pi(1) = 2\pi$$

Step 4: Sum the total distance.
Combine the straightaways and the corner drifts to get the final run length. The total distance covered by the dog in one round is just the sum of our two parts.
$$\text{Total} = 19 + 2\pi$$

Final Answer: $19+2\pi$

The Setup: In $\triangle ABC$, $D$ is the midpoint of $BC$, and $AM$ is the altitude on $BC$. With side ratios $AB:BC:CA = 2:4:3$, we must find the length ratio $BM:AD$.

Step 1: Assign algebraic lengths to the sides.
Let the sides be $c = AB = 2x$, $a = BC = 4x$, and $b = CA = 3x$.
Since $D$ is the midpoint of $BC$, $BD = \frac{BC}{2} = 2x$.

Step 2: Calculate the length of $BM$.
Use the Law of Cosines to find $\cos B$ in $\triangle ABC$:

$$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4x)^2 + (2x)^2 - (3x)^2}{2(4x)(2x)}$$

$$\cos B = \frac{16x^2 + 4x^2 - 9x^2}{16x^2} = \frac{11x^2}{16x^2} = \frac{11}{16}$$

In the right-angled $\triangle ABM$, the segment $BM$ represents the adjacent side to angle $B$. Thus, $BM = c \cdot \cos B$:

$$BM = 2x \left(\frac{11}{16}\right) = \frac{11x}{8}$$

Step 3: Calculate the length of the median $AD$.
Apply Apollonius's Theorem to the median $AD$:

$$AB^2 + AC^2 = 2(AD^2 + BD^2)$$

$$(2x)^2 + (3x)^2 = 2(AD^2 + (2x)^2)$$

$$4x^2 + 9x^2 = 2AD^2 + 8x^2$$

$$13x^2 - 8x^2 = 2AD^2$$

$$5x^2 = 2AD^2 \implies AD^2 = \frac{5x^2}{2} \implies AD = \frac{x\sqrt{5}}{\sqrt{2}} = \frac{x\sqrt{10}}{2}$$

Step 4: Compute the ratio $BM : AD$.

$$\text{Ratio} = \frac{\frac{11x}{8}}{\frac{x\sqrt{10}}{2}} = \frac{11}{8} \times \frac{2}{\sqrt{10}} = \frac{22}{8\sqrt{10}} = \frac{11}{4\sqrt{10}}$$

Final Answer: $11:4\sqrt{10}$

The Setup: In a right-angled triangle $ABC$, the hypotenuse $AC$ is $13\text{ cm}$. The segment $DE$ joining the midpoint of the leg $AB$ to the midpoint of the hypotenuse $AC$ measures $6\text{ cm}$. We must find the length of $BC$.

Step 1: Map the geometry using the Midpoint Theorem.
The problem defines a line segment drawn connecting the midpoints ($D$ and $E$) of two sides of a triangle ($AB$ and $AC$).
According to the Triangle Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is perfectly parallel to the third side and exactly half of its length.

$$DE = \frac{1}{2} BC$$

Step 2: Calculate the base.
We are explicitly given that $DE = 6\text{ cm}$.

$$6 = \frac{1}{2} BC \implies BC = 12\text{ cm}$$

*(The hypotenuse measurement of 13 is consistent data: with $BC = 12$ and $AC = 13$, the remaining leg $AB = 5$, forming the standard 5-12-13 Pythagorean triple.)*

Final Answer: $12$ cm

The Setup: An equilateral triangle is partially overlapped by a circle whose diameter is one of its sides. We must calculate the area of the geometric intersection of the two figures.

Step 1: Establish the geometry of the intersection.
Let the equilateral triangle be $ABC$ with side length $k$. The circle has diameter $AB$, meaning its center $O$ is the midpoint of $AB$, and its radius is $k/2$.
The circle intersects side $AC$ at a point $D$ and side $BC$ at a point $E$.
According to Thales's Theorem, any angle inscribed in a semicircle is a right angle. Since $AB$ is the diameter, $\angle ADB = 90^\circ$ and $\angle AEB = 90^\circ$.
Thus, $BD$ and $AE$ are altitudes of the equilateral triangle $ABC$. In an equilateral triangle, altitudes also act as medians, meaning $D$ and $E$ exactly bisect sides $AC$ and $BC$.

Step 2: Partition the target area.
The portion of the triangle lying inside the circle is the shape bounded by segments $AD$, $EB$, $AB$, and the circular arc $DE$.
We can decompose this bounded region by connecting the center $O$ to points $D$ and $E$.
The area consists of three non-overlapping geometric shapes: $\triangle AOD$, $\triangle BOE$, and the circular sector $DOE$.

Step 3: Calculate the component areas.
Because $O$ is the midpoint of $AB$ and $D$ is the midpoint of $AC$, $AD = k/2$ and $AO = k/2$. Since $D$ is on the circle, $OD = k/2$. Thus, $\triangle AOD$ is equilateral with side length $k/2$.
$$\text{Area}(\triangle AOD) = \frac{\sqrt{3}}{4} \left(\frac{k}{2}\right)^2 = \frac{\sqrt{3}}{16} k^2$$
By identical symmetry, $\triangle BOE$ is also equilateral with the same area.
Because $\angle AOD = 60^\circ$ and $\angle BOE = 60^\circ$, the central angle of the sector $DOE$ is $180^\circ - 60^\circ - 60^\circ = 60^\circ$.
$$\text{Area}(\text{Sector } DOE) = \frac{60^\circ}{360^\circ} \times \pi \left(\frac{k}{2}\right)^2 = \frac{1}{6} \times \frac{\pi k^2}{4} = \frac{\pi}{24} k^2$$

Step 4: Sum the components to find total area.
$$\text{Total Area} = 2 \left( \frac{\sqrt{3}}{16} k^2 \right) + \frac{\pi}{24} k^2$$
$$\text{Total Area} = \frac{\sqrt{3}}{8} k^2 + \frac{\pi}{24} k^2 = \left( \frac{3\sqrt{3}}{24} + \frac{\pi}{24} \right) k^2 = \frac{3\sqrt{3} + \pi}{24} k^2$$

Final Answer: $\left(3\sqrt{3}+\pi\right)\dfrac{k^{2}}{24}$

The Setup: We are hunting for lattice points strictly inside the triangle. The boundary lines are the axes ($x=0, y=0$) and the hypotenuse ($x+y=31$). To stay *interior*, our coordinates must be strictly positive integers that sum to less than 31. Boundary points fail the vibe check and get instantly disqualified.
Step 1: Set the Constraints
We need integers $x > 0$ and $y > 0$ such that $x + y < 31$.
This means the absolute max limit for the sum is $x + y \le 30$.
Step 2: Sum the Rows
If $x = 1$, $y$ can range from $1$ to $29$ (giving us 29 valid points).
If $x = 2$, $y$ can range from $1$ to $28$ (28 points).
This sequence cascades linearly all the way down to $x = 29$, where $y$ can only be $1$ (1 point).
Step 3: Calculate the Total
We just need to sum the first 29 positive integers. We plug this into the classic arithmetic series sum formula $S = \frac{n(n+1)}{2}$.

$$S = \frac{29(30)}{2}$$

$$S = 29 \times 15 = 435$$

Final Answer: 435