Loci and Triangle Centers
Maya and Jordan live on opposite sides of town. They want to meet at a pizza place that is the same distance from both of their houses. Meanwhile, Jordan's dog is tied to a stake with a 5-foot leash and can reach any point 5 feet or less from the stake. Where can the pizza place be? Where can the dog go? Each question asks for a locus: every point that follows a rule.
What Is a Locus?
The simplest locus is the one you already know. The points at a distance $r$ from a fixed point $O$ form the circle with center $O$ and radius $r$. Try it: drag point P and leave "footprints" wherever $OP$ equals the distance you want.
The dog on the leash can reach every point with distance from the stake at most $5$ ft. That is the circle and everything inside it. We write the condition as $d \le 5$. If the leash is pulled taut, the dog is on the circle itself, where $d = 5$.
A young man trains his dog in an open field. The dog is attached to an extendable leash of $5$ feet maximum length, anchored to a stake driven into the ground. Draw and…
A young man trains his dog in an open field. The dog is attached to an extendable leash of $5$ feet maximum length, anchored to a stake driven into the ground. Draw and describe the locus that represents the maximum area the dog can move in if it keeps the leash always taut.
Points at a Fixed Distance from a Line
The distance from a point to a line is always measured perpendicular to the line, because that is the shortest path. Now ask: which points are exactly $d$ away from a line $\ell$?
So if a city bans building within $15$ ft of a bike lane, the edges of the banned zone are two parallel lines, one on each side of the lane. The banned zone is the strip between them.
A new straight bike lane has been designed passing through the center of a city. For safety reasons, no buildings are allowed to be constructed within $15$ feet of this…
A new straight bike lane has been designed passing through the center of a city. For safety reasons, no buildings are allowed to be constructed within $15$ feet of this lane. Describe the locus that delimits the prohibited zone on both sides of the bike lane and draw a simple diagram.
The Perpendicular Bisector
Drag P so that $PA = PB$. What shape do your footprints make?
Why is it perpendicular? Let $M$ be the midpoint of $\overline{AB}$ and let $P$ be any point on the perpendicular through $M$. Triangles $PMA$ and $PMB$ share the side $\overline{PM}$, both have a right angle at $M$, and $AM = BM$. So the triangles are congruent (SAS), and $PA = PB$.
That solves Maya and Jordan's problem. The pizza place can be anywhere on the perpendicular bisector of the segment joining their houses.
Two towns, Town A and Town B, are located at a distance of $8$ miles from each other. The authorities want to build a bus stop that is exactly at the same distance from…
Two towns, Town A and Town B, are located at a distance of $8$ miles from each other. The authorities want to build a bus stop that is exactly at the same distance from both towns. On which line or locus must they locate the stop? Represent it graphically in your notebook.
The Angle Bisector
Remember to measure each distance perpendicular to the side. In the interactive, the colored segments are those perpendiculars.
When two lines cross, they form four angles, so they have two angle bisectors. Those two bisectors are perpendicular to each other. A point equidistant from both lines can be on either one.
In a museum, two walls form an angle of $70^\circ$. A display case is to be installed so that it is always at the same distance from both walls. Draw the angle and trace…
In a museum, two walls form an angle of $70^\circ$. A display case is to be installed so that it is always at the same distance from both walls. Draw the angle and trace the locus where the objects in the display case could be placed. What is the technical name for this line?
Combining Two Conditions
Many problems have two rules at once: "equidistant from $A$ and $B$ and 5 feet from $A$." The answer is the set of points that satisfy both. Draw each locus, then find where the two drawings intersect.
In a school playground, two Wi-Fi antennas are located at points $A$ and $B$, separated by a distance of $12$ feet. A student is looking for a place to sit where they…
In a school playground, two Wi-Fi antennas are located at points $A$ and $B$, separated by a distance of $12$ feet. A student is looking for a place to sit where they receive the same signal strength from both antennas.
- a) Describe what locus is formed by all the points that are at the same distance from $A$ and from $B$.
- b) If the student also wants to be exactly $10$ feet from antenna $A$, in how many possible points in the playground could they sit? Explain what geometric figures you would have to draw to find them.
Two straight streets in a neighborhood, Main Street and Second Street, intersect forming an angle of $60^\circ$. We want to place a trash can that meets two conditions:…
Two straight streets in a neighborhood, Main Street and Second Street, intersect forming an angle of $60^\circ$. We want to place a trash can that meets two conditions: it must be at the same distance from both streets and exactly $5$ feet from the intersection point of the streets.
- a) Define the locus that meets the first condition.
- b) Use the concept of locus to explain how you would determine the exact position of the trash can. How many possible solutions are there in total?
In a soccer practice session, a ball is placed at a point that is equidistant from the two sidelines (which are parallel to each other). If the width of the field is…
In a soccer practice session, a ball is placed at a point that is equidistant from the two sidelines (which are parallel to each other). If the width of the field is $60$ feet:
- a) Describe the locus where the ball can be.
- b) If the coach adds a restriction and says that the ball must be $20$ feet from a specific corner flag, draw a diagram of the situation and identify the points that meet both conditions.
Lines and Circles in Action
Two ideas help when a circle meets a line: the Pythagorean Theorem (to measure a stretch of road or a distance) and the 30-60-90 triangle (to place a circle tangent to two lines).
A circle cutting a line
The closest point of a line to the center is the foot of the perpendicular. That perpendicular, a radius to a crossing point, and the stretch of line between them form a right triangle.
A circle tangent to two lines
A circle that touches both lines has the same distance from its center to each line, namely its radius. So the center must be on an angle bisector of the two lines.
Two crossing lines have two angle bisectors, and on each bisector a center can be on either side of $V$, so there are really four such circles. Usually a problem tells you which one it means.
A lighthouse emits a signal that is only received at a distance of less than $10\text{ mi}$. A boat moves along a straight coastline $c$ which is at a distance of…
A lighthouse emits a signal that is only received at a distance of less than $10\text{ mi}$. A boat moves along a straight coastline $c$ which is at a distance of $8\text{ mi}$ from the lighthouse at its closest point.
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Describe the locus of the points on the coast that receive the lighthouse signal.
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Calculate the length of the stretch of coast where the boat will be within the range of the lighthouse.
Consider two intersecting lines $r$ and $s$ that cross forming an angle of $60^\circ$. We want to draw a circle of radius $10\text{ cm}$ that is tangent to both lines…
Consider two intersecting lines $r$ and $s$ that cross forming an angle of $60^\circ$. We want to draw a circle of radius $10\text{ cm}$ that is tangent to both lines and lies inside the $60^\circ$ angle shown in the figure.
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Explain on which line the center of this circle must be located.
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Calculate at what distance from the intersection point of the two lines the center of the circle must be placed. (Hint: Joining the center to the point of tangency forms a right triangle; use the properties of triangles with angles of $30^\circ$ and $60^\circ$).
Loci on the Coordinate Plane
On a map with coordinates, you can find a perpendicular bisector exactly with algebra. You need three tools:
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Midpoint of $(x_1, y_1)$ and $(x_2, y_2)$: $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$.
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Slope $m = \dfrac{y_2 - y_1}{x_2 - x_1}$. Perpendicular lines have slopes that are negative reciprocals (their product is $-1$). A line with slope $\tfrac{1}{2}$ is perpendicular to one with slope $-2$.
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Point-slope form $y - y_1 = m(x - x_1)$.
A logistics company wants to locate a central warehouse to serve three stores located at points $A(2, 6)$, $B(2, 0)$ and $C(10, 0)$ (units in miles). To minimize…
A logistics company wants to locate a central warehouse to serve three stores located at points $A(2, 6)$, $B(2, 0)$ and $C(10, 0)$ (units in miles). To minimize transport costs, the warehouse must be at the same distance from all three stores.
- a) Determine the coordinates of the point where the warehouse must be located by studying the perpendicular bisectors.
- b) What is the name of the locus of points that are at a distance of $5$ miles from point $B$? Would store $C$ be inside that zone?
A mountain rescue team receives signals from two lost hikers, located at points $A$ and $B$. The command center knows that the hikers are exactly at the same distance…
A mountain rescue team receives signals from two lost hikers, located at points $A$ and $B$. The command center knows that the hikers are exactly at the same distance from each other, but the rescue helicopter must follow a straight trajectory represented by a secondary road $r$. If on a coordinate map the points are $A(1, 6)$ and $B(5, 2)$, and the road follows the direction of the line passing through $(0, 0)$ and $(8, 8)$, determine by means of locus reasoning the exact point where the helicopter will cross the line where the hikers could be found.
The Four Triangle Centers
In every triangle, certain lines always meet in a single point. We say they are concurrent. Two kinds of segments will help us:
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A median goes from a vertex to the midpoint of the opposite side.
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An altitude goes from a vertex perpendicular to the line containing the opposite side.
| Center | Where these meet | Key property | What it is good for |
|---|---|---|---|
| Circumcenter | Perpendicular bisectors | Equidistant from the 3 vertices | Center of the circle through the vertices (circumscribed circle). Place a cell tower so three towns get equal coverage. |
| Incenter | Angle bisectors | Equidistant from the 3 sides | Center of the circle touching the sides (inscribed circle). Put a fountain as far as possible from the edges of a triangular park. |
| Centroid | Medians | Divides each median in a $2 : 1$ ratio | Center of gravity: a triangular cardboard balances on a pencil tip here. |
| Orthocenter | Altitudes | (Lies on the Euler line, see below) | Appears in many geometry proofs. |
Identify the triangle center described in each case: a) It is the point that is equidistant from the three vertices of the triangle. b) It is the point where the three…
Identify the triangle center described in each case:
- a) It is the point that is equidistant from the three vertices of the triangle.
- b) It is the point where the three altitudes of a triangle intersect.
- c) It is the center of the inscribed circle of the triangle (the one that touches its three sides).
- d) It is the center of gravity of the triangle and the point where the medians intersect.
The Circumcenter
Why do the three perpendicular bisectors meet at one point? A point on the bisector of $\overline{AB}$ has $PA = PB$. A point on the bisector of $\overline{BC}$ has $PB = PC$. Where those two bisectors cross, both are true, so $PA = PB = PC$. That point is then also equidistant from $A$ and $C$, so it lies on the third bisector too.
Since this point is the same distance $R$ from all three vertices, the circle with center there and radius $R$ passes through $A$, $B$, and $C$. It is the circumscribed circle, and $R$ is the circumradius.
In a right triangle the circumcenter lands on the midpoint of the hypotenuse, so $R$ is half the hypotenuse.
Draw a triangle in your notebook whose sides measure $6$ in, $8$ in and $10$ in. Draw its three perpendicular bisectors and mark the point where they intersect. Check,…
Draw a triangle in your notebook whose sides measure $6$ in, $8$ in and $10$ in. Draw its three perpendicular bisectors and mark the point where they intersect. Check, using a compass, that this point is the center of a circle that passes through the three vertices. What is the name of this point?
A town council wants to install a street light in a small triangular plaza whose vertices are located at points $P$, $Q$ and $R$. They have decided that the street light…
A town council wants to install a street light in a small triangular plaza whose vertices are located at points $P$, $Q$ and $R$. They have decided that the street light must be at the same distance from the three corners (vertices) of the plaza so that the lighting is balanced.
- a) What is the name of the notable point of the triangle where they must install the street light?
- b) Which notable lines of the triangle is it necessary to draw to locate this point? Draw it schematically.
A telephone antenna is to be installed to serve three villages located at the vertices of a triangle whose sides measure $13\text{ mi}$, $14\text{ mi}$ and…
A telephone antenna is to be installed to serve three villages located at the vertices of a triangle whose sides measure $13\text{ mi}$, $14\text{ mi}$ and $15\text{ mi}$. The chief engineer decides that the antenna must be at the circumcenter of the triangle so that the distance to the three villages is the same. If the area of the triangle is $84\text{ mi}^2$, calculate the coverage radius the antenna must have (the distance from the circumcenter to any of the vertices) knowing that in any triangle the radius of the circumscribed circle satisfies the relationship $R = \frac{a \cdot b \cdot c}{4 \cdot \text{Area}}$.
The Incenter
The same reasoning works for angle bisectors. A point on the bisector of angle $A$ is equidistant from the two sides at $A$. A point on the bisector of angle $B$ is equidistant from the two sides at $B$. Where those two bisectors cross, the point is the same distance from all three sides, so the third bisector passes through it too.
That common distance $r$ is the radius of the inscribed circle (or "incircle"), which touches each side at one point. The radius to each touching point is perpendicular to that side, so the distances from the incenter to the three sides are all equal to $r$.
Here is a handy way to find $r$. Join the incenter to the three vertices. This cuts the triangle into three smaller triangles, each with height $r$ and with the sides of the big triangle as bases. Adding their areas: $$A = \tfrac{1}{2}ar + \tfrac{1}{2}br + \tfrac{1}{2}cr = \tfrac{1}{2}(a + b + c)\,r.$$
A park has a triangular shape. A fountain is to be installed so that the distance from the fountain to each of the three paths (sides of the park) is the same. a) Which…
A park has a triangular shape. A fountain is to be installed so that the distance from the fountain to each of the three paths (sides of the park) is the same.
- a) Which special lines of the triangle would you have to draw to find the location?
- b) What is the resulting point called?
A graphic designer is creating a logo based on a scalene triangle. They want to inscribe a circle inside the triangle so that it is tangent to all three sides. a) What…
A graphic designer is creating a logo based on a scalene triangle. They want to inscribe a circle inside the triangle so that it is tangent to all three sides.
- a) What notable point of the triangle must be the center of this circle?
- b) If the distance from this center to one of the sides is $3$ in, what will be the distance to the other two sides? Justify your answer based on the definition of this locus.
An architect designs a triangular plaza whose sides measure $30$, $40$ and $50$ feet. Inside, he wants to place a fountain that is at the same distance from the three…
An architect designs a triangular plaza whose sides measure $30$, $40$ and $50$ feet. Inside, he wants to place a fountain that is at the same distance from the three streets (the three sides of the triangle).
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What is the name of the locus that determines the position of the fountain?
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Calculate the distance from the fountain to each of the streets. (Hint: The area of a triangle can also be calculated as the product of the semiperimeter and the radius of the inscribed circle, $A = s \cdot r$).
The Centroid
The three medians of a triangle meet at the centroid, the triangle's center of gravity. If you cut a triangle from cardboard, you can balance it on a pencil tip placed at this point.
A board game has been designed where a counter must be placed at the "center of gravity" of a triangular piece of cardboard so that it balances on the point of a pencil.…
A board game has been designed where a counter must be placed at the "center of gravity" of a triangular piece of cardboard so that it balances on the point of a pencil. The sides of the triangle measure $15$ in, $20$ in and $25$ in.
- a) What name is given to this balance point in geometry?
- b) If we draw one of the medians of this triangle, in what proportion does the balance point divide this median? Express the relationship between the distance from the point to the vertex and the distance from the point to the opposite side.
The Euler Line and Special Triangles
You can check this in the triangle figure from earlier: the circumcenter (blue), the centroid (orange), and the orthocenter (pink) line up.
In special triangles, the centers pile up:
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Equilateral triangle: all four centers coincide in a single point, because every median is also an altitude, a perpendicular bisector, and an angle bisector.
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Isosceles triangle: all four centers lie on the axis of symmetry, the line from the apex to the middle of the base. For an isosceles triangle, that axis is the median, the altitude, the angle bisector, and the perpendicular bisector of the base all in one.
Is it possible for the circumcenter, the incenter, the centroid, and the orthocenter to coincide at the same point in a triangle? If so, state what type of triangle it…
Is it possible for the circumcenter, the incenter, the centroid, and the orthocenter to coincide at the same point in a triangle? If so, state what type of triangle it must be and draw it, marking this point.
In an isosceles triangle $ABC$, with $AB = AC$, we know that the centroid (point where the medians intersect), the orthocenter (point where the altitudes intersect) and…
In an isosceles triangle $ABC$, with $AB = AC$, we know that the centroid (point where the medians intersect), the orthocenter (point where the altitudes intersect) and the circumcenter (point where the perpendicular bisectors intersect) are all aligned on the same straight line.
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What is the name of this straight line that contains these notable points?
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If the base $BC$ measures $12\text{ cm}$ and the altitude corresponding to that base measures $9\text{ cm}$, determine the position of the centroid with respect to the base, justifying your answer with the property of the locus of the medians.
The Locus for a Right Angle
Carla stands in a park. Ashley and Brian are $10$ ft apart, and the angle between Carla's lines of sight to them is $90^\circ$. Where can Carla be?
Why is the angle exactly $90^\circ$? Let $M$ be the center, the midpoint of $\overline{AB}$, and let $P$ be on the circle. Then $MA = MB = MP$ (all radii). Triangle $MAP$ is isosceles, so its base angles at $A$ and $P$ are equal. Call each one $\alpha$. Triangle $MBP$ is also isosceles, so its base angles at $B$ and $P$ are equal. Call each one $\beta$. Now triangle $ABP$ has angles $\alpha$, $\beta$, and $\alpha + \beta$ (at $P$). Its angle sum is $\alpha + \beta + (\alpha + \beta) = 180^\circ$, so $\alpha + \beta = 90^\circ$. That is the angle at $P$.
The reverse is also true: if $P$ is inside the circle, the angle is bigger than $90^\circ$. If $P$ is outside the circle, it is smaller than $90^\circ$. Only the circle gives exactly $90^\circ$.
The endpoints $A$ and $B$ are left out because at those points there is no angle $APB$ to measure.
Three friends, Ashley, Brian, and Carla, are located in a park forming a triangle. Ashley and Brian are $10$ feet apart. Carla observes that she is at a point such that…
Three friends, Ashley, Brian, and Carla, are located in a park forming a triangle. Ashley and Brian are $10$ feet apart. Carla observes that she is at a point such that the angle formed by her lines of sight to Ashley and Brian is always $90^\circ$.
- a) What is the name of the locus where Carla is located?
- b) If Carla moves while always maintaining this angle of $90^\circ$ with respect to her friends, what geometric figure would her trajectory describe?
The "arc capable" of an angle $\alpha$ on a segment $AB$ is the locus of points from which the segment subtends the same angle $\alpha$. If we have a segment $AB$ of…
The "arc capable" of an angle $\alpha$ on a segment $AB$ is the locus of points from which the segment subtends the same angle $\alpha$. If we have a segment $AB$ of length $8\text{ cm}$, draw and describe the locus for an angle of $90^\circ$. Justify, based on the properties of inscribed angles, why this locus is a circle of diameter $AB$ and calculate its area.
An Ellipse: Two Fixed Points
Imagine a gardener who puts two stakes in the ground, ties a loop of string around them, and pulls the string tight with a pencil. As the pencil moves, the sum of its distances to the two stakes stays the same, because the string has a fixed length.
Put the center at the origin with the foci on a horizontal axis. Then:
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$a$ = half of the constant sum (the semi-major axis).
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$c$ = half of the distance between the foci.
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$b$ = the distance from the center to the top of the ellipse (the semi-minor axis).
At the top point, the two distances to the foci are equal, so each is $\tfrac{1}{2}$ of the sum, which is $a$. That makes a right triangle with legs $b$ and $c$ and hypotenuse $a$, so $b^2 + c^2 = a^2$.
In an amusement park, a new "suspended flight" ride follows a trajectory such that the sum of its distances to two fixed control towers, $F_1$ and $F_2$, is always…
In an amusement park, a new "suspended flight" ride follows a trajectory such that the sum of its distances to two fixed control towers, $F_1$ and $F_2$, is always $50\text{ meters}$. The towers are separated by a distance of $30\text{ meters}$.
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Identify which locus this is.
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If we place the center of the trajectory at the origin $(0,0)$, what will be the maximum height the ride will reach with respect to the axis joining the two towers? (Hint: use the Pythagorean theorem with the semi-axes of the figure).
Back to the start: Maya and Jordan's pizza place can be anywhere on the perpendicular bisector of the segment between their houses. A display case that must be the same distance from two walls goes on the angle bisector, and Jordan's dog can reach the disk of radius $5$ ft centered at the stake.