Is such a triangle possible?

Question

a triangle angle bac is 45 degrees . side bc is 4 units . altitude from point a is 4 units. Apart from a right angle triangle where altitude becomes side ac is another such triangle possible??

Answer 1

Let $d$ be the foot perpendicular drawn from vertex $a$ to the side $bc$ then

in right $\triangle adc$ $$\angle dac=\angle acd=45^\circ$$$$\iff ad=dc=4$$$$\color{red}{dc=4}\tag 1$$ since the foot of perpendicular $d$ lies between the points $b$ & $c$ on the side $bc$ hence, $$dc<bc=4$$ $\color{red}{dc<4}\tag 2$ from (1) & (2) there is contradiction.

hence such a triangle is not possible.

However, the given triangle is possible if either $bc>4$ or altitude to the side $bc$ is less than $4$ for given angle of $45^\circ$

Answer 2

If you can use trigonometry and the formula for the tangent of the sum then looking at two right triangles ahc and ahb formed by the altitude ah and the base bc split by the point h (with bh+hc=4). You have with bh=x:

1=tan(pi/4)= (tan(hac)+tan(hab))/1-tan(hac)tan(hab)=(x/4+(4-x)/4)(1-(x/4)(4-x)/4)= 16/(16-x(4-x)) thus x(4-x)=0 and either b=h or c=h, i.e the right triangle is the only answer.

A different approach is to see that the altitude and the base are the same, thus the area of the triangle is 8 (4*4/2) independent on the position of the base. Move the little triangle ahc : flip it and glue it to the triangle ahb, along the intervals ah from the triangle ahc to ab from the triangle ahb. Comparing the angles the flipped image of the interval ac will form 45 degree with the old altitude ah. The interval ah as altitude is shorter than ab, and the angle cha is smaller than the external angle b of the triangle ahb. So the flipped part will fit well inside the right isoceles triangle based on the original altitude ah and a 4 units extension of hb, in addition the flipped interval ac is shorter than the hypotenuse of the isoceles as it sees the right angle cha from a shorter distance. But the triangle bac and the isoceles were supposed to be equal in area. Impossible unless there is nothing to flip or we flip the whole triangle.

Answer 3

Given any line segment, $bc$, and a point $a$ on one side of it.

• The locus of constant $\angle bac$ is a circular arc.
• The locus of constant altitude from $a$ to $bc$ is a line parallel to $bc$.

Since a circle arc intersect a line at at most 2 points, there are at most two solutions for $a$.

From your figure, it is clear there are two places for $a$ which fit your requirement. On one of them, $\angle acb = 90^\circ$ and on the other one, $\angle abc = 90^\circ$.

This means all solutions are right angled triangles.

Answer 4

An aspect of the "Inscribed Angle Theorem" indicates that all the points $A$ that make an angle of a given size with points $B$ and $C$ lie on an arc of a circle with $\overline{BC}$ as a chord. (The possible angles $\angle BAC$ are said to be "inscribed" in the circle; hence, the name of the theorem.) The $45^\circ$-angle case is shown here:

(BTW: I like to think of this diagram as the top view of a mouse poking its $45^\circ$ angle nose through a hole in the wall (represented by $\overline{BC}$ ... the rest of the mouse is on the other side of the "hole", where we don't see it). The arc gives the boundary of the area the mouse can reach in its quest for cheese. Obviously, mice with wider noses can't reach as far, and mice with narrower noses can poke-out further; but each mouse has its nosey adventures restricted by some circular arc.)

As the diagram indicates, there are exactly two locations for $A$ that are at distance $4$ from $\overline{BC}$, and each of these leads to a right triangle $\triangle ABC$.

Note also: There are at most two locations for $A$ at any distance you care to name, in general: two locations, if the distance is short enough; no locations, if the distance is too far; or precisely one location, if the distance is just right to graze the "top" of the circle.)